What Is It
To fully understand and describe a projectile’s motion – objects moving in
two dimensions, let us first consider objects moving in one dimension (horizontally
or vertically).
This module will focus on the motion of objects described to be moving at a
constant velocity (having an acceleration of 0 m/s2) or moving at a constant
acceleration. Objects that are moving at a constant acceleration are said to be in a
Uniformly Accelerated Motion (UAM). In other words, the acceleration does not
change; it is uniform throughout.
In the activity, you related distance and time. In computing the slope, you
divided distance by time which is actually the speed of the can. These quantities are
essential in the derivation of formulas that will solve problems relating to uniformly
accelerated motion.
Table 2. Summary of Uniformly Accelerated Motion (UAM) Formulae
Uniformly Accelerated Motion
Formulae
1. vf = at + vi
2. d = vit + at2
2
3. d = vf + vi
t
2
4. Vf2 = vi2 + 2ad
where,
vf
vi
a
t
d
= final velocity
= initial velocity
= acceleration
= time
= displacement
SI unit
m/s
m/s
m/s2
s
m
Uniformly Accelerated Motion (UAM): Horizontal Dimension
If a body maintains a constant change in its velocity in a given time
interval along a straight line, then the body is said to have a uniform
acceleration.
Consider an airplane on a runway preparing for takeoff.
velocity
time
position
0m
5m/s
10m/s
15m/s
1s
2s
3s
5m
10m
20m
30m
45m
Figure 1. An airplane preparing for take-off.
What have you noticed with the velocity of the airplane in the figure above? In the
first second, the velocity is 5m/s, in the next time it becomes 10m/s, and on the
third second it becomes 15m/s. Notice that the velocity of the airplane as it runs on
the runway is increasing. When the velocity changes every second, it means that
the airplane is accelerating. How much is the change in velocity for every second? If
you try calculating for the change in velocity from 1s to 2s, that is 10m/s – 5m/s =
5m/s. The change in velocity from 2s to 3s is 15m/s – 10m/s = 5m/s. Notice that
the change in velocity for every second is 5m/s, it is not changing, it remains the
same or uniform for every second. The figure shows that the airplane is moving
faster at a constant acceleration. This motion is referred to as Uniformly Accelerated
Motion (UAM).
Let’s try solving problems!
Sample Problem1
An airplane from rest accelerates on a runway at 5.50 m/s 2 for 20.25s until
it finally takes off the ground. What is the distance covered before takeoff?
Note: If the airplane started from rest, it means to say that the initial velocity is
zero.
Solution:
1. Write the given or what information is given in the problem.
Given:
Find:
a = 5.50 m/s2
d=?
t = 20.25 s
vi = 0 m/s
2. Next, find out which formula to use in solving the problem. Use the formula
checklist table below to help you find out which formula to use.
Note: Eliminate the formula that does not contain the variable that you need to find.
In this case, we are looking for d (displacement), so eliminate the formula that does
not contain the variable d.
3.
Solve the problem using the chosen formula.
d = vit + at2
2
2
d = (0 m/s)(20.25 s) + (5.50 m/s2) (20.25 s)2
2
2
d = 0 + (5.50 m/s ) (410.0625 s2)
2
d = 1128 m
Sample Problem 2
A motorcyle from rest accelerates uniformly over a time of 3.25 seconds and
covers a distance of 15 m. What is the acceleration of the motorcycle?
Given:
Find:
d = vit + at2
vi = 0 m/s
a=?
21
t = 3.25 s 2
5
m
=
(0
m/s)
(3.25 s)+ a (3.25 s)2
d = 15 m
2
15 m = 0 + a (10.5625 s2)
2
15 m = a (5.28 s2)
a = 15 m
5.28 s2
a = 2.8 m/s2
Try solving this…
A car accelerates to a speed of 15 m/s over 200 m distance.
Determine the acceleration (assume uniform) of the car.
Uniformly Accelerated Motion (UAM): Vertical Dimension
You have been introduced to the concept of gravity in Grade 8. You have
learned that gravity acts on all objects by pulling it towards the center of the Earth.
So, on Earth when you throw something up, it will go down. Things thrown upward
always fall at a constant acceleration (ag) which has a magnitude of 9.8 m/s2. This
means that the velocity of an object in free fall changes by 9.8 m/s every second of
fall.
Consider a stone dropped from a cliff as shown in Figure 2. For equal time
interval, the distance travelled increases quadratically.
0 m/s
-9.8 m/s
-19.6 m/s
-29.4 m/s
-39.2 m/s
Figure 2. Motion of the stone dropped from a hill.
Sample Problem 1
Ben is playing with his ball on top of a building, but the ball fell and hits the
ground after 2.6 seconds, what is the final velocity of the ball just before it hits the
ground and how high is the building?
Note: The initial velocity of an object that is falling or dropped from a height will
always be equal to 0 m/s, since the object started from rest.
Replace the variable d (displacement) with h (height) since we are referring
to the position of an object from top to bottom.
Given:
ag = -9.8 m/s2
assume vi = 0 m/s
t = 2.6 s
Find:
vf = ?
h=?
vf = vi + agt
vf = 0 + (-9.8 m/s2)(2.6 s)
vf = -25 m/s
Acceleration = 1.62
Height = 2m
initial Velocity = 0
h = vit + agt2
2
-h = [ (0 m/s)(2.6 s) ] + (-9.8 m/s2)(2.6 s)2
2
-h = 0 + (-9.8 m/s2)(6.76 s2)
2
-h = -66.248m
2
h = 33 m
Try solving this…
The acceleration of gravity on the moon is 1.62 m/s2. If a ball is dropped on
the moon from a height of 2 m. Determine the time for the ball to fall to the surface
of the moon.
Motion in Two Dimensions
Have you seen a soccer player kick a ball? Or a basketball player shoots a ball
into the ring? What have you noticed with the path the balls travel? If you noticed,
the balls travel a curve path. The curve naturally happens when an object, called a
projectile, moves in two dimensions – having both horizontal and vertical motion
components, acted by gravity only. In physics, this is called projectile motion.
Understanding the Physics of projectile helps players enhance their game
skills and experience.
Projectile motion can happen in two situations – projectiles launched
horizontally and projectiles launched at an angle. This module will describe motion
of projectiles launched horizontally.
Projectiles Launched Horizontally
A projectile launched horizontally has no
initial vertical velocity (vix = 0). Thus, its vertical
motion is similar to that of a dropped object.
Projectiles are only acted upon by gravity.
Therefore projectiles launched horizontally only
have downward acceleration (ay), which is
equivalent to -9.8 m/s2 and has no horizontal
acceleration (ax = 0).
Projectiles have horizontal (x) and vertical
(y) components that are independent of each other.
Thus, we will recreate the Uniformly Accelerated
Motion Formulae according to the horizontal and
vertical components of a projectile.
Figure 3. Velocity component
vector diagram for
horizontally-fired projectile.
Table 4. Horizontal and Vertical Component Formulae for Projectiles
Launched Horizontally
Uniformly
Accelerated
Motion
Formulae
Horizontal (x)
component
Formulae
ax = 0
vf x= axt + vix
1. vf = at + vi
2. d = vit + at2
2
3. d = vf + vi
2
2
2
4. Vf = vi +
2ad
t
Vertical (y)
component
Formulae
viy = 0
vf x = vix
X = vixt + axt2
X=vixt
2
X = vfx + vix_
t
2
Vfx2 = vix2 + 2axx
Vfx2 = vix2
ax =horizontal acceleration
vfx = final horizontal velocity
vix = initial horizontal velocity
x = horizontal displacement
vf y= ayt + viy
ay t
vf y =
y= viy t + ay t2
2
y = vfy_
t
y =ayt2
2
2
Vfy2
Vfy2
= viy2 +2ayy
= 2ayy
ay = vertical acceleration (ag)
vfy = final vertical velocity
vix = initial vertical velocity
y = vertical displacement
Sample Problem 1
A marble is thrown horizontally from a table top with a velocity of 1.50 m/s.
The marble falls 0.70 m away from the table’s edge.
A) How high is the lab table?
B) What is the marble’s velocity just before it hits the floor?
Before you can find the height of the lab table, you must determine first the
time it took the marble to reach the ground.
Given/Find:
Horizontal (x)
component
vix = 1.50 m/s
x = 0.70 m
Vertical (y)
component
ay = -9.8 m/s2
y=?
vfy = ?
t=?
Note: You can solve for time using equations from Horizontal or Vertical Component
depending on the given. In this problem, we cannot solve for time using the Y
component since the variables are lacking. If you look at the X Component, Vix and
X are given; therefore, we can use these variables in solving for time.
Solve for time using this formula: X =vixt
X = vixt
0.70 m = (1.50 m/s) t
t = _0.70 m_
1.50 m/s
t = 0.47 s
Since we already have the value for time = 0.47s, we can now solve for
y (vertical displacement) or the height of the lab table. We can use this formula in
solving for y: y = ayt2
2
y = ayt2
2
2) (0.47 s)2
-y
=
(-9.8
m/s
y takes a negative
2
sign since we are
2) (0.2209 s2)
-y
=
(-9.8
m/s
dealing with
2
downward
-y
=
-2.16482
m
displacement
2
-y = -1.08 m
y = 1.08 m
We can also solve for vfy using our value for time = 0.47 s. We can use
this formula : vf y = ayt
vf y = ayt
vf y = (-9.8 m/s2) (0.47 s)
vf y = 4.61 m/s
What’s More
33
Solve the Problems
Directions: Solve the following problems. Show complete solution.
1. Maria dropped a coin in the wishing well and it hits the bottom after 3 seconds.
Determine the final velocity of the coin as it hits the bottom of the well and find out
the height of the wishing well.
Given/Find
Formula/e
Solution (4 points)
(3 points)
(2 points)
Given:
ag=________
vi = _______
t = ________
Find:
vf = ?
h=?
a._______________
b. ______________
2. An airplane from rest accelerates on a runway at 6.5 m/s 2 for 15 s until it finally
takes off the ground. What is the distance covered before takeoff?
Given/Find
Formula/e
Solution (2 points)
(3 points)
(1 point)
Given:
vi = ________
a = ________
t = _________
Find:
d=?
a._______________
3. A motorcycle from rest accelerates uniformly over a time of 5 seconds and covers a
distance of 20 m. Determine the acceleration of the motorcycle.
Given/Find
(4 points)
Given:
vi = ________
t = _________
d = ________
Formula/e
(1 point)
Solution (2 points)
a._______________
Find:
___________
4. A cannonball is launched horizontally from the top of a mountain with a velocity of
3.5 m/s. The cannonball falls 50 m away from the edge of the mountain.
a) How high is the mountain?
b) What is the cannonball’s velocity just before it hits the ground?
Given/Find:
Horizontal (x) component (2 pts) Vertical (y) component (3 pts)
vix = _______
ay = ________
Find:
x =________
________
________
t=?
Solution: (6 pts)
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