3.3 Projectile Motion
The motion of an object that is subjected to a constant,
downward acceleration is termed projectile motion. We already
covered 1D projectile motion in chapter 2 – there, objects were
either dropped or thrown vertically. The constant acceleration
(a = -g) dictated the position of the object over time.
Now, we will allow the object to move in two dimensions (why
not 3?) As we saw in the last lecture, we can break up this 2D
problem into two 1D problems, for the horizontal (x) and vertical
(y) directions. By combining these two solutions, we can
determine the path over which the object travels. The relevant
equations are repeated here (originally from slide 9 of the last
lecture):
1
2
𝑥 = 𝑥0 + 𝑣𝑥0 𝑡 + 𝑎𝑥 𝑡
2
1
𝑦 = 𝑦0 + 𝑣𝑦0 𝑡 + 𝑎𝑦 𝑡 2
2
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𝑣𝑥 = 𝑣𝑥0 + 𝑎𝑥 𝑡
𝑣𝑦 = 𝑣𝑦0 + 𝑎𝑦 𝑡
Day 6 – May 16 – WBL 3.3-3.4
Slide 1
3.3 Projectile Motion
Since the acceleration is downward, we can state that ay = -g and
ax = 0. In other words, the horizontal motion is defined by zero
acceleration (constant velocity), and the vertical motion is
defined by constant acceleration.
Horizontal Projection
We begin with the case that the
projectile’s initial velocity is
horizontal.
Since there is no horizontal
acceleration, we can say that at all
times, vx = vx0. Furthermore, this
means that the horizontal position is
a linear function of time: x = x0 + vx0t.
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Slide 2
3.3 Projectile Motion
While the projectile is traveling at a constant horizontal velocity,
it is also traveling vertically. However, since there is a non-zero
vertical acceleration, the vertical velocity is not constant. This
causes the projectile to move in a curved path.
The figure to the right shows two objects that
are released at the same time – the red ball is
dropped from rest while the yellow ball is
projected horizontally. At any given time, both
balls have dropped the same vertical
distance. The vertical motion and horizontal
motion are “uncoupled” (they don’t depend
on each other). Both balls will land at exactly
the same time.
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Slide 3
3.3 Projectile Motion
Projections at Arbitrary Angles
The most general case of projectile motion is one in which the
object is launched at an arbitrary angle θ relative to the
horizontal. It has initial velocity components 𝑣𝑥0 = 𝑣0 cos 𝜃 and
𝑣𝑦0 = 𝑣0 sin 𝜃.
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Slide 4
3.3 Projectile Motion
Since there is no horizontal acceleration, the x-component of
velocity is constant. It will always equal 𝒗𝟎 cos 𝜽.
However, since there is vertical acceleration (ay = -g), the ycomponent of velocity varies with time, according to the 1D
kinematics equations from chapter 2. In summary,
𝑣𝑥 = 𝑣𝑥0 = 𝑣0 cos 𝜃
𝑣𝑦 = 𝑣𝑦0 − 𝑔𝑡 = 𝑣0 sin 𝜃 − 𝑔𝑡
As the figure on the previous slide indicates, the projectile will
land with the same speed 𝑣0 at which it was launched (although
the vertical component of velocity will be reversed in sign). The
angle at which it lands will be −𝜃. This assumes that the landing
height is the same as the launch height.
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Slide 5
3.3 Projectile Motion
If we place the origin at the launch point, then 𝑥0 , 𝑦0 = (0,0).
In this case, the kinematic equations from chapter 2 (with ax = 0)
produce
𝑥 = 𝑣𝑥0 𝑡 = 𝑣0 cos 𝜃 𝑡
1 2
1 2
𝑦 = 𝑣𝑦0 𝑡 − 𝑔𝑡 = (𝑣0 sin 𝜃)𝑡 − 𝑔𝑡
2
2
In the case that the launch point is not the origin, we can simply
replace 𝑥 with 𝑥 − 𝑥0 and 𝑦 with 𝑦 − 𝑦0 .
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Slide 6
3.3 Projectile Motion
At this point, the text proclaims that “the curve described by
these equations…is called a parabola”. Let’s prove that, since it’s
not too hard.
The previous equations describe x and y as functions of time.
However, the figure on slide 4 shows the trajectory of the
projectile, which is actually y as a function of x. Nothing is
plotted as a function of time (although the position of the object
is shown at 7 particular times during the projectile’s flight).
To find y(x), we follow two steps:
1. Invert the x(t) equation to find t(x)
2. Substitute this into the y(t) equation to convert it to y(x)
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Slide 7
3.3 Projectile Motion
Following these steps, we find that
𝑥
𝑡=
𝑣0 cos 𝜃
And therefore
2
𝑣0 sin 𝜃
1
𝑥
𝑔
2
𝑦=
𝑥− 𝑔
= tan 𝜃 𝑥 − 2
𝑥
𝑣0 cos 𝜃
2
𝑣0 cos 𝜃
2𝑣0 cos 2 𝜃
Since 𝜃, 𝑣0 , and 𝑔 are all constants, this function has the form
𝑦 = 𝑎𝑥 − 𝑏𝑥 2 , which describes a parabola.
Again, if the launch position is (𝑥0 , 𝑦0 ) rather than (0,0), then we
simply replace 𝑥 with 𝑥 − 𝑥0 and 𝑦 with 𝑦 − 𝑦0 .
PC141 Intersession 2013
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Slide 8
Problem #1: Thrown Football
WBL LP 3.9 (modified)
A football is thrown on a long pass. Compared to the
ball’s initial horizontal velocity component, the magnitude
of the velocity at the highest point is…
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A
greater
B
less
C
the same
Day 6 – May 16 – WBL 3.3-3.4
Slide 9
3.3 Projectile Motion
A very important aspect of projectile motion is the range, R. This
is the horizontal distance traveled when the object returns to its
original height.
The text describes one method of calculating R on p. 83. Here’s
another method: by the definition of R, we’re simply looking for
the value of x that results in y = 0. One solution is x = 0.
However, this just reproduces the launch point. The more useful
solution (which requires a bit of algebra and trigonometry) is
𝑣02 sin 2𝜃
𝑅=
𝑔
The range is a function of the launch conditions (𝑣0 and 𝜃) as
well as the magnitude of the gravitational acceleration (g).
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Slide 10
3.3 Projectile Motion
Now that we know where a projectile will land, it’s natural to
ask: is there an optimum launch angle, which will maximize the
range? This is obviously quite important in areas such as
athletics and artillery fire.
Since 𝑅 =
𝑣02 sin 2𝜃
,
𝑔
if 𝑣0 and 𝑔 are constant, we can maximize R
by maximizing sin 2𝜃. This occurs when 𝜃 = 45°, and produces
𝑣02
𝑅max =
𝑔
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Slide 11
3.3 Projectile Motion
In real life, projectiles are subject to air resistance. Incorporating
this factor into an analysis of projectile motion is beyond the
scope of PC141. If we were to do so, we would find that the
optimum launch angle is somewhat less than 45°.
For the rest of this course, unless otherwise specified, assume
that there is no air resistance.
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Slide 12
Problem #2: Kicked Football
The figure shows three paths for a football kicked from ground level. Note that all
three paths have the same maximum height, but they have different ranges.
a)
b)
c)
d)
Rank them in terms of time of flight (longest to shortest)
Rank them in terms of initial vertical velocity component (fastest to slowest)
Rank them in terms of initial horizontal velocity component (fastest to slowest)
Rank them in terms of initial speed (fastest to slowest)
Solution: In class
PC141 Intersession 2013
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Slide 13
Problem #3: Long Jump
In the 1991 World Track and Field Championships, Mike Powell jumped a distance
of 8.95 m (breaking a 23-year-old world record by 5 cm). Assume that Powell’s
speed on takeoff was 9.5 m/s.
a) How much less was Powell’s range than the maximum possible range for an
object launched at the same speed?
b) What is the greatest source of the discrepancy in a)?
c) What would Powell’s distance have been if he had run just a bit faster, with a
speed on takeoff of 9.6 m/s (and the same initial angle as before)?
Solution: In class
PC141 Intersession 2013
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Slide 14
Problem #4: Flying Fish
Fish use various techniques to avoid their predators. The California flying fish
propels itself out of the water using its tail, at a typical speed of 30 km/h.
a) If this fish left the water at a 45° angle, how far could it travel through the air?
b) Suppose a nearby fisherman measured the distance as 180 m. How could this
be explained?
Solution: In class
PC141 Intersession 2013
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Slide 15
Problem #5: Volleyball
A volleyball court is 18.0 m long (end-to-end) , and the top of the net is 2.24 m
above the ground (for women’s competition). Using a jump serve, a player strikes
the ball at a point that is 3.00 m above the ground, and 8.00 m away from the net.
The initial motion of the ball is horizontal.
a) What is the minimum initial velocity if the ball is to clear the net?
b) What is the maximum initial velocity of the ball is to land in bounds on the
other side of the net?
Solution: In class
PC141 Intersession 2013
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Slide 16
3.4 Relative Velocity
In chapter 2, we learned that an object’s position is measured
relative to an arbitrary origin.
As it turns out, any measurement of an object’s velocity is also
dependent on the velocity of the observer. As an example, you
may think you’re jogging at 5 m/s, but that’s only relative to a
reference frame that is pinned to the Earth. Since the Earth is
spinning about its axis, and rotating about the sun (which itself is
moving relative to the center of the Milky Way galaxy), an alien
who is observing you from far away would claim that you’re
moving much faster than 5 m/s.
In most cases in PC141, the “pinned to the Earth” reference
frame is appropriate. However, sometimes we need to deal with
velocities relative to each other.
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Slide 17
3.4 Relative Velocity
Relative velocities are indicated
using multiple subscripts. For
example, v𝐵𝐴 means “velocity of
object B relative to observer A”.
When the velocities of A and B
are measured relative to a
common reference frame, then
v𝐵𝐴 = v𝐵 − v𝐴
(note the order of the subscripts).
In the figure, velocities are shown relative to the Earth. Since A
km
km
is stationary, we have v𝐵𝐴 = 90
x − 0 = 90
x and
h
h
km
km
v𝐶𝐴 = −60
x − 0 = −60
x
h
h
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Slide 18
3.4 Relative Velocity
In this figure, there is no physical distinction in comparison to the
last figure. However, now we want to measure velocities relative
to car B (that is, relative to a reference frame that is pinned to
car B rather than to the Earth). In this case, we need to find v𝐴𝐵
and v𝐶𝐵 . Using the same subscript ordering as before, we have
km
km
v𝐴𝐵 = v𝐴 -v𝐵 = 0 − 90
x = −90
x
h
h
In other words, the driver of car B sees car A as traveling at 90
km/h in the negative direction.
Similarly, v𝐶𝐵 = v𝐶 -v𝐵
km
= −150
x
h
More examples can be found on
pp. 90-92 of the text.
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Slide 19
3.4 Relative Velocity
Sometimes, the various velocities in a physical problem do not
have a common reference point. Here’s a 1D example from the
text (the results are still valid in 2D and 3D).
Consider a moving walkway (w) in an airport that moves relative
to the ground (g) with velocity 𝑣𝑤𝑔 . A passenger (p) walks on it
with a velocity relative to the walkway of 𝑣𝑝𝑤 . The passenger’s
velocity relative to the ground is
𝑣𝑝𝑔 = 𝑣𝑝𝑤 + 𝑣𝑤𝑔
Again, the ordering of the subscripts is important. The “inner”
subscripts in the sum are identical, and the “outer” subscripts
are those which appear in the left side. Page 90 of the text
contains numerical examples.
PC141 Intersession 2013
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Slide 20
Problem #6: Oncoming Traffic
WBL LP 3.11
You are traveling in a car on a straight, level road going 70
km/h. A car coming toward you appears to be traveling with a
speed of 130 km/h. How fast is the other car actually going?
A
130 km/h
B
60 km/h
C
D
70 km/h
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80 km/h
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Slide 21
Problem #7: Airplane with Crosswind
WBL Ex 3.84 (modified)
An airplane is flying at 150 km/h (its speed in still air) in a direction such that with a
wind of 60 km/h blowing from east to west, the airplane travels in a straight line
southward.
a) What must be the plane’s heading (direction) for it to fly directly south?
b) If the plane has to go 200 km in the southward direction, how long does it take?
Solution: In class
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Day 6 – May 16 – WBL 3.3-3.4
Slide 22