1. Parabolic or Projectile Motion
2-dimentional motion
Parabolic or Projectile Motion

2. Projectile Motion
A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola. 2

3. Motion in Two Dimensions3

4. Motion in Two Dimensions
ax = 0
ay = -g 4

5. Motion in Two Dimensions
Ignoring air resistance, the horizontal component of a projectile's acceleration
(A) is zero.
(B) remains a non-zero constant.
(C) continuously increases.
(D) continuously decreases. 5

6. Solving Problems Involving Projectile Motion
Read the problem carefully, and choose the object(s) you are going to analyze.
Draw a diagram.
Choose an origin and a coordinate system.
Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.
Examine the x and y motions separately. 6

7. Solving Problems Involving Projectile Motion
6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point.
7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them. 7

8. Problem
A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below.
a) How high was the cliff? y d v
b) How far from its base did the diver hit the water? 8

9. First convert cm to m, 24.3 cm = .243 m
A marble rolls off the edge of a table with a height of m and strikes the floor at a distance of 24.3 cm from the edge of the table. Calculate the initial velocity of the marble.
First convert cm to m, 24.3 cm = .243 m
ax = 0
dx = 0.243m (how far the ball traveled)
viy = 0
g = m/s2
dy = (a negative direction)
Find vix
g = m/s2 m
0.243 m

10. dy = - 12.10 (a negative direction) Find dx
An archer stands on the wall of a castle and fires an arrow from a height of m above the ground. If the archer fires an arrow parallel to the ground with an initial horizontal velocity of 11.0 m/s, how far will the arrow travel horizontally before hitting the ground?
ax = 0
vfx = 0
vix = 11.0 m/s
viy = 0
g = m/s2
dy = (a negative direction)
Find dx
g = m/s2 m

11. dy = 1/2gDt2 Dt = 2dy = 2 (-12.10 m) = 1.57 s -9.81 m/s2 vix = dx Dt
First find how long the arrow will be in the air
dy = 1/2gDt2
Dt = 2dy = 2 ( m) = 1.57 s
-9.81 m/s2
Next, we can make use of the initial velocity in the horizontal and the change in time to find the displacement
vix = dx Dt
dx = vix Dt = (11.0 m/s)(1.57 s) = 17.3 m

12. An airplane is flying a practice bombing run by dropping bombs on an old shed. The plane is flying horizontally with a speed of 185 m/s. It releases a bomb when it is 593 m away from the shed, and it scores a direct hit. How high was the plane flying when it dropped the bomb?
ax = 0
vfx = 0
vix = 185 m/s
dx = 593m
viy = 0
g = m/s2
Find dy
593 m

13. vix = dx dx = vix Dt Dt = dx 593m = 3.21 s Dt vix 185 m/s
First we start by using the initial horizontal velocity and the horizontal displacement to determine how long the bomb was in the air
vix = dx dx = vix Dt Dt = dx m = 3.21 s
Dt vix m/s
Now that we know how long it takes the bomb to fall, we can calculate the bomb’s vertical displacement.
dy = 1/2gDt2 = (0.5)(- 9.81m/s2)(3.21 s)2 = m
(a negative direction)

14. dy = - 18.6 m (a negative direction) Find vix
A car launches horizontally off the edge of a cliff with a height of 18.6 m. It strikes the ground below at a distance of 98.4 m away from the edge of the cliff. How fast was the car going when it flew off the edge of the cliff?
ax = 0
vfx = 0
dx = 98.4 m
viy = 0
g = m/s2
dy = m (a negative direction)
Find vix m
98.4 m

15. dy = - 18.6 m (a negative direction) Find vix
A car launches horizontally off the edge of a cliff with a height of 18.6 m. It strikes the ground below at a distance of 98.4 m away from the edge of the cliff. How fast was the car going when it flew off the edge of the cliff?
ax = 0
vfx = 0
dx = 98.4 m
viy = 0
g = m/s2
dy = m (a negative direction)
Find vix m
98.4 m

16. dy = 1/2gDt2 Dt = 2dy = 2(-18.6 m) = 1.95s g -9.81 m/s
First we start by using the vertical displacement and the acceleration due to gravity to determine how long the car was in the air
dy = 1/2gDt Dt = 2dy = 2(-18.6 m) = 1.95s
g m/s
Now that we know how long it takes the car to fall, we can calculate the car’s initial velocity.
Vix = m = m/s
1.95 s