Physics
HSC Course
Stage 6
Space
Part 2: Projectile motion
Contents
Introduction ............................................................................... 2
Projectile motion........................................................................ 4
The trajectory – two separate parts.....................................................5
Determining various quantities ................................................ 15
x and y components ...........................................................................15
Velocity of the projectile .....................................................................16
Maximum height and trip time ...........................................................18
Range of the trajectory.......................................................................20
Escape velocity ....................................................................... 25
Summary................................................................................. 29
Suggested answers................................................................. 31
Exercises – Part 2 ................................................................... 35
Part 2: Projectile motion
1
Introduction
In the previous unit of work you learned about the nature of
gravitational fields. You will now begin to learn about escaping the
Earth’s gravitational field in order to reach space. Your focus will be on
examining the motion of an object projected into the air, but not
propelled after launch like a rocket. This type of motion is called
projectile motion.
Before beginning this part you must have already studied certain
concepts. In particular
you must be able to rdescribe
r
r r and calculate
r Dv v – u
r
r
velocity ( v = D ) and acceleration ( a =
).
=
Dt
t
t
These were covered when you studied the module Moving about in the
preliminary physics course.
In Part 2 you will be given the opportunities to learn to:
•
describe the trajectory of an object undergoing projectile motion
within the Earth’s gravitational field in terms of horizontal and
vertical components
•
describe Galileo’s analysis of projectile motion
•
explain the concept of escape velocity in terms of the:
•
–
gravitational constant
–
mass and radius of the planet
discuss Newton’s analysis of escape velocity.
In Part 2 you will be given opportunities to:
•
2
solve problems and analyse information to calculate the actual
velocity of a projectile from its horizontal and vertical components
Space
•
solve problems and analyse information using:
r
r
r
v = u + at
r
r
r
v 2 = u 2 + 2as
r
r
1r
s = ut + at 2
2
in relation to projectile motion
•
perform a first-hand investigation, gather secondary information and
analyse data to describe factors, such as initial and final velocity,
maximum height reached, range, time of flight of a projectile, and
quantitatively calculate each for a range of situations by using
simulations, data loggers and computer analysis.
Extract from Physics Stage 6 Syllabus © Board of Studies NSW, 1999.
The original and most up-to-date version of this document can be found on the
Board’s website at http://www.boardofstudies.nsw.edu.au.
Solving projectile motion problems is a skill acquired through practice.
In this part you will learn the methods you need to do this. However, just
reading about the methods is not enough. You must attempt all the
problems assigned. Check your answers as you progress and if you are
incorrect look at the solutions provided to find out why, then attempt the
problem again. This practice is important and is the best way to learn
that skill.
Part 2: Projectile motion
3
Projectile motion
A projectile is an object that is projected (that means thrown, dropped or
launched) into the air, but not propelled as is a rocket. This includes a
ball being thrown, a football being kicked, a golf ball being struck, a
bullet being fired, or cargo being dropped from a plane.
Four hundred years ago, Galileo Galilei stated that all masses, large or
small, fall at the same rate. It is difficult, here on Earth, to convince
people of this and he conducted experiments to prove his point. He is
reputed to have dropped different objects from the leaning tower of Pisa.
However, this wasn’t conclusive so he moved on to more sophisticated
experiments. He had two major obstacles to overcome. The first is that
gravity makes things happen too quickly to get conclusive results with
crude equipment. The second problem was that resistance from the air
affected large surface area objects, slowing their acceleration.
He eventually overcame these difficulties by designing highly polished
ramped tracks, down which he would roll balls of different masses.
This reduced the effective acceleration. The lower rate of acceleration
was less affected by air resistance and was easier to measure. Therefore,
he was able to show that the rate of acceleration due to gravity is the
same for all masses. You saw in the previous part that this rate has an
average value of 9.80 ms2 and that this acceleration was a direct result of
the gravitational force of attraction between the Earth and any mass
within its gravitational field. This is the force we call ‘weight’.
Throughout a projectile’s motion following its launch, there are just two
forces acting upon it – the weight force down and air resistance acting
against its velocity at any instant. To simplify your treatment of
projectile motion ignore air resistance so that, for your purposes, weight
is the only force affecting a projectile’s motion.
4
Space
The trajectory – two separate parts
A launched projectile, in the absence of any air resistance, will follow a
parabolic path until it strikes the Earth. This path is called the trajectory
of a projectile. The effect of air resistance is to reduce the maximum
height and horizontal distance the projectile could otherwise achieve.
The diagram below shows the trajectory of a ball thrown with and
without air resistance. The exact effect of air resistance upon a trajectory
depends upon many variables (such as speed of the projectile and air as
well as the shape, size and spin of the projectile) so ignore it in the
calculations that follow.
parabolic trajectory
of a projectile
without air resistance
trajectory with
air resistance is
reduced in height
and range
Air resistance affects the trajectory of a projectile.
To analyse the trajectory of a projectile you need to make a very
important observation: the motion of a projectile can be regarded as two
separate and independent motions superimposed upon each other.
•
The horizontal motion, is not subject to any forces, and therefore
experiences no acceleration.
•
The vertical motion, is subject to the vertical weight force, and
therefore experiences the downwards acceleration due to gravity.
(Recall that there is only one force acting – the vertical weight.)
Part 2: Projectile motion
5
In which direction does gravity always work? Did you say, down?
Because the two motions are perpendicular, they are independent.
Therefore, you can treat them, and analyse them, separately. The
diagram below shows both of these motions within an x-y frame of
reference. This frame will become important.
y
uniformly
accelerated
vertical
motion
non-accelerated
horizontal
motion
x
The motion of a projectile can be analysed as two separate motions.
Acceleration equations
At this point you should take a few moments to recall some of the work
from the preliminary course topic called Moving about. In that module
you studied uniformly accelerated motion, that is, motion that is subject
to acceleration that is unchanging. This acceleration was defined by the
following equation.
6
Space
r r r
r Dv v – u
a=
=
Dt
t
where
r
a = acceleration, in ms-2
r
Dv = change in velocity, in ms-1
r
v = final velocity, in ms-1
r
u = final velocity, in ms-1
t = time taken, in s
Although this is the defining equation, you will more often see this
equation in the following rearranged form:
r r r
v = u + at
Using this equation, two more equations describing accelerated motion
can be derived. (You don’t need to know the derivations, but you can
look them up in a good physics reference if you are curious.) These
equations are:
rr
v 2 = u 2 + 2ar
where
r
r = displacement covered, in m
r r 1r 2
r = ut + at
2
As you examine each of the horizontal and vertical motions, you are
going to change these last three equations to more specific equations that
better suit your purposes.
The horizontal motion
The horizontal portion of a projectile’s motion is like the motion of a
horizontally sliding object. Can you imagine an object sliding in a
horizontal direction without friction? It’s difficult, isn’t it? But, that’s
only because we are so used to dealing with friction from personal
experiences. However, if you have ever played air hockey, then you
have seen the way a puck can slide sideways slowly and, without any
acceleration at all, progress its way across the table. This is
non-accelerated, uniform velocity motion, just like the horizontal part of
projectile motion.
Non-accelerated uniform velocity.
Part 2: Projectile motion
7
The motion is not accelerated because the only force present, the
downward weight force, is at right angles to the horizontal motion, and
therefore does not influence it. In order to modify the three equations
above you need only note that acceleration equals zero. Also, you are
going to use different symbols for displacement and velocity, to indicate
motion in the x direction (horizontal). The changes are:
r
a =0
r
a = 0
r
v = vx
r
u = ux
r
r = Dx (displacement = change of position on the x - axis)
If these changes are substituted into the three equations then we get the
following:
r
v x = u x (that is, horizontal velocity is uniform)
v 2x = u 2x (this really just says the same as first equation)
Dx = u x t
Sample problem 1
Let’s start to use these formulas right away. A rifle with a muzzle
velocity (the speed the bullet comes out of the barrel) of 450 ms-1 is fired
level at the horizon. Determine:
a)
how fast the bullet is travelling 0.3 seconds after firing
b) how far it has travelled horizontally in that time.
Solution
a)
b)
r
Our first formula tells us that v x = u x , that is, the final velocity
equals the initial velocity over any time period. In other words, the
horizontal velocity is the same all the way through the motion.
Therefore, the velocity after 0.3 s is still 450 ms-1 (horizontally).
Dx = u x t
= 450 ¥ 0.3
= 135 m
That is, after 0.3 s the bullet has travelled 135 m.
Try this next problem out for yourself. These motion problems are quite
simple, however solutions are provided at the end of this booklet.
8
Space
An air hockey puck is pushed so that it glides along its table at 0.15 ms-1. If
the table is 1.2 m long, determine:
a)
how long the puck takes to travel the length of the table
_____________________________________________________
_____________________________________________________
_____________________________________________________
b) its velocity when it gets there.
_____________________________________________________
_____________________________________________________
_____________________________________________________
Check your answers.
The range of the trajectory
There is one more thing to point out regarding the horizontal motion – it
doesn’t go on forever. Eventually, the falling projectile will strike the
Earth and stop moving. The horizontal displacement then has a
maximum value, called the range of the trajectory. Also, the time t has a
maximum value, that can be called the trip time. In order to calculate
what the trip time will be it is necessary to examine the vertical motion of
a projectile’s motion.
Sample problem 2
A stone is thrown horizontally at 8.0 ms-1. If it takes 0.5 s to fall to the
ground, how far horizontally will it have travelled in this time?
Solution
It is helpful to list the data as you read a question. It will help you to
develop the skill of interpreting numerical physics problems
ux = 8.0 ms-1, t = 0.5 s, Dx = ?
Dx = u x t
= 8.0 ¥ 0.5
= 4.0 ms-1
That is, the trajectory of the stone will have a range of 4.0 m.
Part 2: Projectile motion
9
Now it is your turn to try to determine the range of a trajectory. A tennis
-1
ball is struck horizontally at 15.0 ms . It is low to the ground so that it takes
just 0.3 s to strike the ground. How far does the ball travel horizontally
before it bounces for the first time?
_________________________________________________________
_________________________________________________________
_________________________________________________________
Check your answer.
You should now attempt Exercises 2.1 and 2.2.
The vertical motion
Recall that the only force acting on a projectile during its motion is the
downward weight (ignoring air resistance). This force will affect the
vertical portion of the motion of a projectile. According to Isaac
Newton’s second law of motion, a net force acting on a mass will cause
an acceleration. The acceleration in this case is, of course, acceleration
r
due to gravity g . This was discussed in some detail in the previous part
where you deduced an average value of around 9.8 ms–2.
As a result, the vertical portion of projectile motion
is uniformly accelerated motion where the rate of
acceleration is 9.8 ms-2.
In this regard it is much like any object thrown
straight up. If thrown up from ground level, an
object will rise up to a peak height, stop
momentarily in the air, then return to the ground,
speeding up as it does so. The second half of the
motion is symmetrical with the first part, so that the
time taken for the fall will equal the time taken for
the rise. Also, the speed with which the object
strikes the ground will equal the speed with which
it was launched (only the direction will be down
instead of up).
This symmetry will be used to solve many
problems later on.
10
Space
You now need to adapt the three acceleration equations for the vertical
motion. To do so you will make the following changes to the variables:
a = ay = 9.8 ms-2 down
v = vy
u = uy
r = Dy (displacement = change of position on the y-axis)
Remember that the vertical direction corresponds to the y-axis in our x-y
frame of reference. Substituting these changes:
vy = uy + a y t
v 2y = u 2y + 2a y Dy
1
Dy = u y t + a y t 2
2
This now gives a set of three equations that can be used specifically for
uniformly accelerated vertical motion.
Sample problem 4
An arrow is fired directly upwards with a velocity of 55 ms-1. Assume
that it is fired from ground level and that there is no air resistance. There
are a number of things about the motion of this arrow that you need to be
able to calculate. Each is modelled below.
a) How fast is the arrow moving when it returns to the ground?
Solution: By the symmetry of the motion, you can say that the arrow
will have a velocity of 55 ms-1 down.
b) What is the time of flight of the arrow?
Solution: A useful strategy to solve this problem is to focus on the
arrow’s rise up to its peak height. Assume that ‘up’ is the positive
direction. You can now say that:
uy = 55 ms-1, vy = 0 ms-1, ay = -9.8 ms-2, t = ?
The equation to use is:
vy = uy + a y t
0 = 55 + (–9.8)t
\ t = 5.6 s
That is, the arrow will take 5.6 s to rise to its peak height.
By symmetry, it must take just as long to fall back, so:
trip time = 2 ¥ 5.6 = 11.2 s.
Part 2: Projectile motion
11
In other words, the arrow will take 11.2 s to return to the ground.
You need to be careful about the number of significant figures that
you quote in the answer. The data used had only two significant
figures, the answer should be restricted to the same number of
significant figures. This is because in physics the number of
significant figures implies the accuracy with which a piece of data is
known. If the trip time is stated as 11.2 s you are implying that you
know the trip time more accurately than the data from which it was
calculated. This isn’t possible! Therefore, you should state that the
trip time is approximately 11 s (two significant figures).
c)
What is the maximum height reached by the arrow?
Solution: Once again consider just the rise up to the peak of
trajectory. Note that at the peak the arrow will stop momentarily, so
its velocity at this point is zero. The data available is:
uy = 55 ms-1
vy = 0 ms-1
ay = -9.8 ms-2
Dy = ?
Selecting an equation that has these four variables:
v 2 = u 2y + 2a y Dy
0 2 = 552 + 2 ¥ ( -9.8)Dy
\ Dy = 154 m ª 150 m
That is, the peak height achieved by the arrow is approximately
150 m. Notice again the restricted number of significant figures in
the answer, in order to conform with the data.
d) What is the velocity of the arrow 7.5 s after firing?
Solution: In this case it is best to consider the whole of the motion
rather than specific segments. The data is:
uy = 55 ms-1
ay = -9.8 ms-2
t = 7.5 s
vy = ?
vy = uy + a y t
= 55 + ( -9.8 ¥ 7.5)
= -18.5 ms-1
That is, the velocity of the arrow after 7.5 s is approximately 19 ms-1
downwards.
12
Space
Sample problem 4 illustrates the basic strategies for solving vertical
uniform-acceleration problems. Note the very useful practice of listing
the available data before stating a problem. This helps you to select the
right equation to use from the start. It’s worth pointing out that in the
HSC exam marks are allocated for working, so you must always show it!
Following are two problems for you to practice on. Attempt each before you
r
-2
check the answers. For each problem, use g = 9.8 ms down.
1
In the first problem, a sandbag is dropped from a balloon that is
stationary, in the air, 84 m above the ground.
a) Calculate how long the sandbag takes to reach the ground.
_________________________________________________
_________________________________________________
_________________________________________________
b) Determine the velocity of the sandbag just before it strikes the
ground.
_________________________________________________
_________________________________________________
_________________________________________________
2
This next problem is a bit harder. A tennis player tosses the ball into
the air to serve It takes 1.1 s between tossing and serving. Assume
that the racquet strikes the ball at the same height from which it was
tossed.
a) Calculate the speed with which the ball was tossed. (Hint:
displacement equals zero for the whole motion.)
_________________________________________________
_________________________________________________
_________________________________________________
b) Calculate the height achieved, above the player’s hand, by the
ball.
_________________________________________________
_________________________________________________
_________________________________________________
Check your answers.
You should now attempt Exercises 2.3 and 2.4.
Part 2: Projectile motion
13
Putting the two parts together
By now you have adapted the three acceleration equations to suit.
It always helps to look for patterns in the information you are trying to
learn, so take a moment to view them into one table. Notice that the
major difference between the equation sets is the absence of acceleration
in the x direction. Note also that the one variable common to both sets is
the time, t.
Horizontal motion
Vertical motion
x direction
y direction
vx = ux
vy = uy + a y t
v 2x = u 2x
v 2y = u 2y + 2a y Dy
Dx = u x t
1
2
Dy = u y t + a y t 2
This is the toolkit of equations that you can call upon to solve problems
that involve both horizontal and vertical motion, that is, whole projectile
motion problems. Even so, some strategy is required to determine
particular quantities. Now you are going to examine the methods you
should use to do this.
14
Space
Determining various quantities
x and y components
Often the first step you will need to do is to resolve the initial velocity of
the projectile into horizontal and vertical components, so that these can
then be used in the various equations. This is a simple application of
trigonometry, as shown by the following diagram.
u
q
uy = u sin q
ux = u cos q
The mathematical expressions you need to use therefore are:
•
the horizontal component, ux = u cos q
•
the vertical component, uy = u sin q , where q is measured from
horizontal.
Sample problem 5
What is the horizontal and vertical components of the initial velocity of a
tennis ball struck at 12.5 ms-1 at 30.0° above horizontal?
Part 2: Projectile motion
15
Solution
The horizontal component, ux = u cos q = 12.5 ¥ cos 30.0 = 10.8 ms-1
The vertical component, uy = u sin q = 12.5 ¥ sin 30.0 = 6.25 ms-1
Here is a problem for you to practice on. Check your answers with those
provided at the end of this unit after you have finished (not before!).
-1
A soccer player kicks the ball off the ground with a velocity of 7.0 ms at
8.0° above horizontal. What are the horizontal and vertical components of
the ball’s velocity?
_________________________________________________________
_________________________________________________________
Check your answer.
Velocity of the projectile
In order to find the velocity of the projectile at other times during its
flight you need to separately calculate the velocity in the x direction and
the y direction, and then add them together using two-dimensional vector
arithmetic. (This is made easier by the two vectors being at right angles
to each other.) The various steps that you need to follow are as follows:
a)
Resolve initial velocity u into components ux and uy.
b) Note that the horizontal velocity is uniform and doesn’t change
(vx = u x )
c)
Consider the vertical motion and calculate the velocity vy after the
specified time.
d) You must now add vx and vy together as shown in the following
diagram:
Vx
q
V
q
16
Vy
or
Vy
V
Vx
Space
Note that the magnitude of v is given by v = v 2x + v 2y (Pythagoras’
theorem) and the angle is given by:
Êu ˆ
q = tan -1 Á y ˜
Ë ux ¯
If you need more information on resolving vectors you may wish to
recall the work on resolving vectors from the module Moving about,
Part 2.
Sample problem 6
A tennis ball is struck, this time at 5.0 ms-1, 55° above horizontal.
What is the velocity of the tennis ball 1.2 s after being struck?
a)
The components of the initial velocity are:
-1
The horizontal component, ux = u cos q = 5.0 ¥ cos 55 = 2.87 ms
The vertical component, uy = u sin q = 5.0 ¥ sin 55 = 4.10 ms-1
b) vx = ux = 2.87 ms-1
c)
In analysing the vertical motion we shall take ‘up’ to be the positive
direction.
v y = u y + a y t = 4.10 + (-9.8 ¥ 1.2) = -7.66 ms-1= 7.66 ms-1 down
d) The final step is to add vx and vy together as shown in the diagram
below.
q
2.87 ms-1
7.66 ms-1
V
Part 2: Projectile motion
17
v = v 2x + v 2y = 2.872 + 7.66 2 = 8.2 ms-1
Êv ˆ
7.66 ˆ
q = tan -1 Á y ˜ = tan -1 Ê
= 69∞
Ë 2.87 ¯
Ë vx ¯
That is, after 1.2 s the velocity of the tennis ball is 8.2 ms-1 at 69° below
horizontal.
Continue now with your practice problem, by returning to the soccer player
used before (‘A soccer player kicks the ball off the ground with a velocity of
7 ms-1 at 8° above horizontal’). Calculate the velocity of the ball 0.05 s after
it was kicked.
_________________________________________________________
_________________________________________________________
_________________________________________________________
_________________________________________________________
Check your answer.
Maximum height and trip time
The strategy for working out these quantities is to consider just the
vertical motion up to the peak. At the peak, the projectile has stopped
moving in the vertical direction so that you can say that vy = 0 in this
portion of the motion. The method then follows these steps.
a)
Resolve the vertical component uy of the initial velocity u.
b) Consider the vertical motion up to the peak.
c)
Note that vy = 0 for this portion of the motion.
d) Select an acceleration equation that enables calculation of unknowns
from the data.
then either
calculate Dy, which is the maximum height.
or
calculate t, which is the time for the projectile to rise up to the peak.
e)
18
Double this time to find the trip time. (Making use of the symmetry
of the motion here, because it takes as long to fall as it does to rise
up to the peak.)
Space
Sample problem 7
Back to the tennis player for this problem. This time the player plays a
half volley off the ground, so that the ball leaves the racquet with a
velocity of 7.2 ms-1 at 36° above horizontal. Calculate the maximum
height achieved by the ball, and the time it takes to bounce for the first
time (that is, the trip time).
Solution
The first step is to calculate uy:
uy = 7.2 sin 36∞ = 4.2 ms-1
Now considering the vertical motion up to the peak, noting that for this
segment, vy = 0. The available data is:
uy = 4.2 ms-1, vy = 0 ms-1, ay = -9.8 ms-2, Dy = ?
The calculation needed to find the maximum height is:
v 2 = u 2y + 2a y Dy
0 2 = 4.2 2 + 2( -9.8)Dy
Dy = 0.9 m
That is, the maximum height achieved by the ball is 0.9 m. To calculate
the trip time you need to find the time to reach the peak. The available
data now is:
uy = 4.2 ms-1, vy = 0 ms-1, ay = -9.8 ms-1, Dy = 0.9 m, t = ?
A suitable calculation to find the time to the peak is:
vy = uy + a y t
0 = 4.2 + ( -9.8)t
\ t = 0.43 s
and hence, trip time = 2t = 2 ¥ 0.43 = 0.86 s
Part 2: Projectile motion
19
D
=
x
x
tu
Returning to the soccer player once again. You have already calculated the
components of the initial velocity and the velocity of the ball just 0.05 s after
being kicked. Your task now is to calculate the maximum height achieved
by the ball and the time it takes to return to the ground.
_________________________________________________________
_________________________________________________________
_________________________________________________________
_________________________________________________________
_________________________________________________________
Check your answer.
Range of the trajectory
Your final strategy is concerned with calculating the maximum
horizontal displacement of a projectile, that is, the range of the trajectory.
In order to make this calculation you will need to follow these steps:
a)
Resolve initial velocity u into components uy and ux.
b) Analyse the vertical motion to find the trip time as shown above.
c)
Consider only the horizontal motion and calculate the range using,
Dx = u x t
Sample problem 8
Back to the tennis shot played in the earlier sample problem. The ball
was struck from ground level at 7.2 ms-1 at 36° above horizontal.
What will be the range of its trajectory?
Solution
The first step, as usual, is to resolve the initial velocity into components.
ux = 7.2 cos 36∞ = 5.8 ms-1
uy = 7.2 sin 36∞ = 4.2 ms-1
Normally you would have to calculate the trip time by analysing the
vertical motion, but this has already been done in the solution to sample
problem 7. You know that the trip time is 0.86 s.
20
Space
The final step then is to analyse the horizontal motion to find the
maximum displacement. This is the displacement that corresponds to the
trip time. The available data is:
ux = 5.8 ms-1, trip time t = 0.86 s, Dx = ?
The required calculation is:
Dx = u x t
= 5.8 ¥ 0.86
= 5.0 m
That is, the tennis ball travelled 5.0 m before bouncing for the first time.
1
Once again it is your turn to try the calculation. Returning to the soccer
player again. In the previous portion of this practice problem you
calculated the time taken for the ball to return to the ground, and you
have already calculated ux for the ball. Use these pieces of information
to calculate how far the ball travels horizontally before it bounces.
_____________________________________________________
_____________________________________________________
2
In this problem, a person stands on the edge of a 50.0 m high cliff
and kicks a ball horizontally out at 12 ms-1 so that it falls into the
valley below. Calculate how far from the base of the cliff the ball
will land. Assume that the cliff is vertical and that there is no air
resistance. (This question is a little different to the others we have
done so far. Think about it a little first and you will realise that it is
actually easier to solve.)
_____________________________________________________
_____________________________________________________
_____________________________________________________
3
The last practice problem is of the more usual sort. A hockey player
strikes the ball, giving it a velocity of 15 ms-1 but unfortunately lifts
the ball off the grass at an angle of 5.0° above horizontal. In hockey,
the ball is not supposed to rise above knee height. If we assume that
this is 0.4 m above the ground, is the hockey player playing an
illegal shot?
_____________________________________________________
_____________________________________________________
_____________________________________________________
Check your answers.
Part 2: Projectile motion
21
You should now attempt exercises 2.5, 2.6, 2.7 and 2.8.
To download computer simulation programs of projectile motion see sites
on the physics website page at http://www.lmpc.edu.au/science
Test your projectile motion analysis skills
In this activity you are going to set up a simple projectile motion, using a
marble as the projectile. Before performing the experiment you will
calculate the range of the projectile’s trajectory and place a cup at this
location to catch the marble. The marble will then be released. Will you
succeed?
You will need:
•
a marble
•
a bench
•
a ruler which has ridges that can act as a track for the marble.
Method:
1
The ruler is to be set up as shown in the diagram.
marble
q
bench
plastic
cup
floor
Experimental set up to calculate the range of a projectile’s trajectory.
22
Space
The upper end of the ruler should be propped up with a few books
and the lower end should be a few centimetres away from the edge
of the bench. The marble will be held at the upper end of the ruler
and then released. It will roll down the ruler, onto the bench, and
then over the edge of the bench. From that point it is a projectile.
2
Measure the height of the upper end of the ruler above the bench,
and record this.
Height of upper end of ruler = ________________ cm
3
Now use trigonometry to calculate the angle the ruler is making with
the bench.
-1
Angle q = sin (height of upper end of ruler / length of ruler)
= _________________ °
4
The component of the acceleration due to gravity directed down the
slope of the ruler can now be calculated.
aslope = g sin q = ______________________ ms-2.
5
Your next step is to consider the acceleration of the marble down the
ruler in order to calculate the velocity of the marble when it reaches
the bench surface.
v = u + a t = _________________________________ ms-1
This velocity will now become the initial horizontal velocity of the
marble when it rolls off the bench and into the air.
6
The next step is to analyse the vertical portion of the projectile
motion in order to determine the trip time.
Measure the height of your bench and record it here:
Dy = _________________ m
Next, calculate the time taken for the marble to fall to the floor:
_____________________________________________________
_____________________________________________________
_____________________________________________________
Therefore, trip time = ____________ s
7
The final step in this analysis is to consider the horizontal motion to
determine the range.
_____________________________________________________
_____________________________________________________
_____________________________________________________
Therefore, the range will be _______________ m
Part 2: Projectile motion
23
8
Now it is time to put your prediction to the test. Place a plastic cup
this distance from the base of the cupboard. If your bench top has a
lip that extends beyond the base of the cupboard you will need to
compensate for this. Finally, place the marble at the top of the ruler
and release it.
Did your marble land in your cup? __________________________
If not, can you suggest reasons why? (It may have fallen short, and
this would indicate a loss of energy.)
______________________________________________________
______________________________________________________
______________________________________________________
24
Space
Escape velocity
Isaac Newton had some interesting thoughts about projectile motion.
In your treatment of projectiles you have been assuming that the ground
is flat but on a larger scale, of course, it is not. Rather, it shows the
curvature of the Earth. Newton reasoned that if projected far enough,
a projectile will start to make its way around the curvature of the Earth
before returning to the ground. If projected fast enough then a projectile
will make it completely around the Earth before returning to its starting
point. He drew sketches to illustrate this idea, that looked like this:
Projectile motion according to Newton.
Newton had no way of testing this idea by experimentation, of course,
but it does lead to a question: what if the projectile had been projected
even faster than this velocity?
The idea was used by Jules Verne in his book From the Earth to the
Moon about two hundred years later. Verne described a spacecraft
launched by cannon. Its speed out of the cannon would be sufficient to
allow the spacecraft to escape the gravity of the Earth and head off
towards the Moon. It sounds impractical, doesn’t it? In fact, this scheme
wouldn’t work, but only for purely practical reasons.
Part 2: Projectile motion
25
The Earth has an atmosphere that causes friction with anything that tries
to pass through it quickly, leading to a heating effect. The speed of a
spacecraft out of such a cannon would be so great that the spacecraft
would burn up almost immediately after hitting the atmosphere.
Additionally, occupants of this spacecraft would never survive the
enormous g force suffered during this launch.
If these practical problems are removed, by assuming that the spacecraft
has no living occupants and the Earth has no atmosphere, then it is quite
possible to calculate just how fast it must be fired directly up so that it
completely escapes the Earth’s gravitational field. This may surprise you
to read this, given that the idea is such an old one. In order to determine
this ‘escape velocity’ you will need to recall some of the ideas about the
Earth’s gravitational field that were discussed in the previous part.
Firstly, recall that a mass has not escaped the Earth’s (or any planet’s)
gravitational field until it is at an infinite distance away (theoretically,
anyway). A more practical term is ‘a very large distance away’. At this
point its gravitational potential energy is zero, and at any point closer it
has a negative potential energy.
Jules Verne’s cannon will be giving the spacecraft kinetic energy (recall
1
that E k = mv 2 ) and if this equals the magnitude of the spacecraft’s
2
gravitational potential energy at the surface of the Earth, then the
spacecraft will have enough energy to leave the Earth’s gravitational
field completely. Therefore:
Ek = Ep
m m
1
mv 2 = G E
2
rE
v=
where
v
2 Gm E
rE
= escape velocity, in ms-1
24
mE = mass of the Earth = 5.974 ¥ 10 kg
rE
= radius of the Earth = 6.378 ¥ 106 m
G
= universal gravitational constant = 6.67 ¥ 10
-11
Nm2kg-2
A more general form of this equation that can be applied to any planet
would be:
Escape velocity =
26
2 Gm planet
rplanet
Space
where
mplanet
= mass of the planet, in kg
rplanet
= radius of the planet, in m
Notice that the escape velocity depends upon just two variables – the
mass of a planet and its radius. Interestingly, it does not depend upon the
mass of the spacecraft or projectile involved. With this formula you can
now calculate what the velocity of the spacecraft fired out of Jules
Verne’s cannon would have to have been:
Escape velocity =
=
2 Gm Earth
rEarth
2(6.67 ¥ 10 -11 )(5.97 ¥ 10 24 )
6.38 ¥ 10 6
= 11 200 ms-1 ª 40 000 kmh -1
That is, the escape velocity of Earth is about 40 000 kmh-1. To put this
into some perspective, an FA/18 jet fighter plane can fly at
approximately twice the speed of sound, or mach two. Expressed in the
same terms, the escape velocity of the Earth is approximately mach 33!
This is a very high velocity and now you can see why, in practice, it
would present several problems.
Sample problem 9
What is the escape velocity of the planet Mercury, given that its mass is
3.6 ¥ 1023 kg and its radius is 2439 km.
Solution
escape velocity =
=
2 Gm Mercury
rMercury
2(6.67 ¥ 10 -11 )(3.6 ¥ 10 23 )
2.439 ¥ 10 6
= 4 437 ms-1 ª 16 000 kmh -1
That is, escape velocity on the planet Mercury is approximately
16 000 kmh-1.
Part 2: Projectile motion
27
Here is a practice problem for you to attempt.
Calculate the escape velocity of the planet Mars. The mass of Mars is
approximately 6.57 ¥ 1023 kg and its diameter is 6 795 km. (Watch out for
the little trick here.)
_________________________________________________________
_________________________________________________________
_________________________________________________________
Check your answer.
Complete Exercises 2.9 and 2.10.
In this part you have learned about the nature of projectile motion and
acquired the skills necessary to solve projectile motion problems.
In addition, you have learned about the concept of escape velocity.
In the next part, you will learn about the orbital motion of satellites.
28
Space
Summary
•
A projectile is any object that is projected into the air but does not
continue to be propelled.
•
The path of a projectile, called its trajectory, has a parabolic shape if
air resistance is ignored. The trajectory can be analysed
mathematically by regarding the horizontal and vertical components
of the motion separately.
•
The horizontal motion of a projectile is constant velocity and can be
analysed using these equations:
vx = ux
v 2x = u 2x
Dx = u x t
•
The vertical motion of a projectile is uniformly accelerated motion
and can be analysed using these equations:
vy = uy + a y t
v 2y = u 2y + 2a y Dy
1
Dy = u y t + a y t 2
2
•
Escape velocity is the velocity with which an object must be
projected vertically in order to completely escape the gravitational
field of a planet.
Escape velocity =
where
G
2 Gm planet
rplanet
= universal gravitational constant
= 6.67 ¥ 10-11 Nm2kg-2
Part 2: Projectile motion
mplanet
= mass of the planet, in kg
rplanet
= radius of the planet, in m
29
30
Space
Suggested answers
The horizontal motion
a)
ux = 0.15 ms-1, x = 1.2 m, t = ?
Dx = u x t
1.2 = 0.15 ¥ t
\ t = 8.0 s
b) vx = ux = 0.15 ms-1
The range of the trajectory
ux = 15.0 ms-1, t = 0.3 s, Dx = ?
Dx = u x t
= 0.15 ¥ 0.3
= 4.5 m
The vertical motion
1
Balloon problem:
a) uy = 0 ms-1, ay = -9.8 m s-2, Dy = -8.4 m, t = ?
1
2
Dy = u y t + a y t 2
1
2
-8.4 = 0 ¥ t + ( -9.8)t 2
\ t = 4.1 s
b) uy = 0 ms-1, ay = -9.8 m s-2, Dy = -8.4 m, vy = ?
v 2y = u 2y + 2a y Dy
= 0 2 + 2 ¥ ( -9.8)( -84)
\ v y = 41 ms-1
Part 2: Projectile motion
31
2
Tennis player problem:
a) Consider the whole motion: t = 1.1 s, a = -9.8 ms-2, Dy = 0 m,
uy = ?
1
2
Dy = u y t + a y t 2
1
2
0 = u y ¥ 1.1 + ( -9.8)(1.1)2
\ u y = 10.5 ms-1
b) Consider now the rise up to the peak: uy = 10.5 ms-1,
a = -9.8 ms-2, vy = 0 ms-1, Dy = ?
t = 1.1 ∏ 2 = 0.55 s
1
2
Dy = u y t + a y t 2
1
2
= 10.5 ¥ 1.1 + ( -9.8)(0.55)2
= 4.3 m
x and y components
The horizontal component, ux = u cos q = 7.0 ¥ cos 8.0 = 6.9 ms-1
The vertical component, uy = u sin q = 7.0 ¥ sin 8.0 = 0.97 ms-1
Velocity of the projectile
Note that vx = ux = 6.9 ms-1
Consider the vertical motion to find vy:
v y = u y + a y t = 0.97 + (-9.8 ¥ 0.05) = 0.48 ms-1
Use vector addition to combine vx and vy together to give v:
v = v 2x + v 2y = 6.92 + 0.482 = 6.9 ms-1
( )
Êv ˆ
0.48
q = tan -1 Á y ˜ = tan -1
= 4o above horizontal
6.9
Ëv ¯
x
32
Space
Maximum height and trip time
Consider the motion up to the peak: uy = 0.97 ms-1, vy = 0 ms-1, ay = -9.8
ms-2, Dy = ? t = ?
v 2y = u 2y + 2a y Dy
0 2 = 0.972 + 2( -9.8)Dy
\ Dy = 0.048 m = 4.8 cm
That is, the maximum height achieved by the soccer ball is 4.8 cm.
Consider the vertical motion to the peak in order to calculate the time
taken:
vy = uy + a y t
0 = 0.97 + 2( -9.8)Dy
\ t = 0.049 s
and hence, trip time = 2t = 2 ¥ 0.049 ª 0.20 s
Range of the trajectory
1
Consider the horizontal motion: ux = 6.9 ms-1, t = 0.20 s, Dx = ?
Dx = u x t
= 6.9 ¥ 0.20
ª 1.4 m
2
Cliff problem. Consider the vertical motion first:
uy = 0 ms-1, ay =-9.8 ms-2, Dy =-50.0 m, t = ?
1
2
Dy = u y t + a y t 2
1
2
-50 = 0 ¥ t + ( -9.8)t 2
\ t = 3.2 s
Consider the horizontal motion to calculate the range:
ux = 12 ms-1, t = 3.2 s, Dx = ?
Dx = u x t
= 12 ¥ 3.2
ª 38 m
Part 2: Projectile motion
33
3
Hockey problem. Consider the vertical motion to the peak:
vy = 0 ms-1, ay = -9.8 ms-2, Dy = ?
uy = u sin q = 15 ¥ sin 5.0 = 1.3 ms-1
v 2y = u 2y + 2a y Dy
0 2 = 1.32 + 2( -9.8)Dy
\ Dy = 0.086 m = 8.6 cm
Therefore, this shot is well below the specified height so it is legal.
Escape velocity
mMars = 6.57 ¥ 1023 kg, rMars = 6.795 ¥ 106 m ∏ 2 = 3.398 ¥ 106 m
escape velocity =
=
2 Gm Mercury
rMercury
2(6.67 ¥ 10 -11 )(3.6 ¥ 10 23 )
3.398 ¥ 10 6
= 5 080 ms-1 ª 18 300 kmh -1
34
Space
Exercises – Part 2
Exercises 2.1 to 2.10
Name: _________________________________
Complete the exercises and return them to your teacher if you are a
distance education school student. If you are an Open Learning Program
TAFE student your teacher will supply you with the answers to these
exercises.
By doing these exercises you should learn whether or not you have
understood the main concepts taught, and achieved the outcomes for this
section of the course. Your teacher will send comments back to you to
help you achieve any outcomes you are not currently achieving.
Data:
r
acceleration due to gravity, g = 9.8 ms-2
Exercise 2.1
a) Describe the shape of the trajectory of a projectile, launched into the
air at an angle to the horizontal.
_____________________________________________________
_____________________________________________________
b) You analyse projectile motion as two separate motion a horizontal
and vertical motion. Why are the two motions independent of each
other?
_____________________________________________________
_____________________________________________________
c) Describe the difference in the nature of the horizontal and vertical
motions.
_____________________________________________________
_____________________________________________________
Part 2: Projectile motion
35
d) What was Galileo’s contribution to projectile motion?
______________________________________________________
______________________________________________________
Exercise 2.2
a) A snooker ball is struck so that it has a velocity of 0.25 ms-1. If the
snooker table is 1.3 m long, how long does this ball take to travel the
full length of the table?
______________________________________________________
______________________________________________________
______________________________________________________
b) A spherical bomb is rolled across a smooth floor. It has a velocity of
2.7 ms-1 and explodes 4.8 s after being released. How far has it
travelled when it explodes?
______________________________________________________
______________________________________________________
______________________________________________________
Exercise 2.3
A science teacher conducts a demonstration of gas pressure. The test
tube containing water was fitted gently with a cork, and then held
vertically in a clamp. It was then heated over a Bunsen burner until the
cork popped out with a velocity of 4.2 ms-1.
a) Calculate the height to which the cork rises.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
b) Calculate the time taken for the cork to rise and fall back to the level
of the mouth of the test tube.
______________________________________________________
______________________________________________________
36
Space
Exercise 2.4
Ernie loves to eat his peanuts by throwing them up in the air and catching
them in his mouth. When he throws them up his hand is level with his
mouth. Each peanut is in the air for 1.4 s.
a) How fast is Ernie throwing the peanuts?
_____________________________________________________
_____________________________________________________
_____________________________________________________
b) How high above his mouth do the peanuts rise?
_____________________________________________________
_____________________________________________________
_____________________________________________________
Exercise 2.5
Scary McLairy, the famous stuntman, is planning his next job. In this
stunt he will stand on the bonnet of a car travelling at 72 kmh-1 as it has a
collision with another car. Just prior to the collision, Scary will jump up
with an initial velocity of 4.9 ms-1. His resultant projectile motion will
catapult him over the collision so that her can land on the boot lid of a
third car. How far away must the third car be, for Scary to successfully
perform his stunt?
4.9 ms-1
72 kmh-1
range = ?
_________________________________________________________
_________________________________________________________
_________________________________________________________
_________________________________________________________
_________________________________________________________
Part 2: Projectile motion
37
Exercise 2.6
When trying to bowl, Tim, a novice ten-pin bowler, slightly ‘lofts’ the
ball – a practice that is frowned upon. This means that he released the
ball to late and lifted it into the air, so that it crashes down upon the
wooden flooring of the lane. If the ball was released with a velocity of
3.5 ms-1 and at an angle of 15° above horizontal, calculate:
a) the x and y components of the initial velocity.
______________________________________________________
______________________________________________________
b) the maximum height achieved by the ball.
______________________________________________________
______________________________________________________
c) the time of flight of the ball.
______________________________________________________
______________________________________________________
d) how far down the lane the ball lands, measured from Tim’s release
point
______________________________________________________
______________________________________________________
e)
all of the above again if the ball was 20% heavier.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
Exercise 2.7
a) A coastal defence cannon is positioned at the top of a 150 m vertical
cliff. It is able to fire a one kilogram shell at a velocity of 120 ms-1.
If fired horizontally out from the cliff, how far out to sea will the
shell land?
______________________________________________________
______________________________________________________
______________________________________________________
38
Space
b) The cannon is now aimed up at an angle of 45° above horizontal.
How far will a shell now travel?
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
Exercise 2.8
A rifle is fired up at an angle of 55° above horizontal. If the initial
velocity of the bullet is 570 ms-1, what will be its velocity 2.5 s after
firing?
_________________________________________________________
_________________________________________________________
_________________________________________________________
Exercise 2.9
a) Upon what variables does the escape velocity from a planet depend?
_____________________________________________________
_____________________________________________________
b) Describe Isaac Newton’s explanation of escape velocity.
_____________________________________________________
_____________________________________________________
Exercise 2.10
a) The planet Venus has a mass of 4.9 ¥ 1024 kg and a radius of
6052 km. Determine its escape velocity.
_____________________________________________________
_____________________________________________________
b) Determine the escape velocity of the planet Pluto, given that its mass
is 1.8 ¥ 1022 kg and its diameter is 2320 km.
_____________________________________________________
_____________________________________________________
Part 2: Projectile motion
39