1. Quiz
1. An object is dropped from a height of 6.5 meters. How long does it take to reach the ground?
2. An object is moving at a constant velocity of 5.7 m/s. How far does the object travel in 1.3 seconds?

2. Independent Vector Analysis
Projectile Motion
Independent Vector Analysis

3. Objectives
Recognize that the vertical and horizontal motions of a projectile are independent.
Relate the height, time in the air and initial vertical

4. Terms Projectile Any object that moves through the air Trajectory
The path a projectile takes through space
Range
The horizontal distance a projectile moves from the launching point

5. What is Projectile Motion?
When a projectile
is given an initial thrust,
and we ignore air resistance,
it moves through the air
with only what acting on it?
Gravity!

6. Bullet Fired vs. Bullet Droped
Mythbusters Youtube
Draw it in your notes
Conclusion – Vertical Motion (in the y-direction) is accelerated because of gravity.

7. Packaged Dropped from Plane
Draw it in your notes.
Click on the red x to drop the package.
During its decent, press pause at several different times.
Where is the package, relative to the plane, as it falls to the ground?
Conclusion – Horizontal Motion (in the x-direction) is constant velocity because
Horizontal acceleration = ax = 0 = zero

8. How Does Projectile Motion Work?
The vertical an horizontal motion of a projectile are independent of each other

9. How Does Projectile Motion Work?
We know that the horizontal motion of a launched ball does not affect its vertical motion.
A projectile launched horizontally has no initial vertical velocity
The vertical and horizontal motion of a projectile are independent of each other
Vi = 0 m/s

10. How Does Projectile Motion Work?
Therefore a horizontally launched projectile has a vertical motion just like an object in freefall.
And horizontal motion just like an object in constant motion.

11. How Does Projectile Motion Work?
Therefore a horizontally launched projectile has a vertical motion just like an object in freefall.
And horizontal motion just like an object in constant motion.

12. Problem Solving Strategies
Motion in two dimensions can be solved by breaking the problem into two interconnected one-dimensional problems.
vf = vi + at
vi a t vf x vi a t vf R y
vf = vi + at

13. Problem Solving Strategies
Projectile motion can be divided into a vertical motion problem and a horizontal motion problem.
v = d/t
The horizontal component can be solved like a constant motion problem
v t d
d = vit + ½ at2
The vertical component can be solved like a freefall problem vi a t vf d

14. Problem Solving Strategies
Both the vertical and horizontal components of a projectile are connected by time (t).
The time for a projectile to hit the ground is the same for the vertical component and the horizontal component. tx
tx = ty = tR tP ty

15. Example 1
A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff?
What is Step 1?
Draw a picture
dx = ?
vx = 5 m/s
tx = ?
vi = 0 m/s
g = -9.8 m/s2
dy = 78.4 m
ty = ?
tP=
ty = tx
vf = ?

16. Example 1 (continued) Horizontal Vertical G: U: E: S:
A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff?
Horizontal Vertical
dy = 78.4 m
vi = 0 m/s
a = -9.8 m/s2
Vx = 5.0 m/s G: U: E: S:
dx = ?
tx = ?
ty = ?
vf = ?
d = vit + ½ at2
d = ½ at2
t = 2d a
t =
2 (78.4 m)
(-9.8m/s2)
t = 4.0 s

17. Example 2 Horizontal Vertical G: U: E: S:
A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff m high. How far from the cliff does the stone hit the ground ?
Horizontal Vertical
dy = 78.4 m
vi = 0 m/s
a = -9.8 m/s2
vx = 5.0 m/s G: U: E: S:
t = 4.0 s
dx = ?
tx = ?
ty = ?
vf = ?
d = vit + ½ at2
d = ½ at2
t =
v = d/t
d = vt 2d a
d = (5.0 m/s)●(4.0 s)
t =
2 (78.4 m)
(-9.8m/s2)
d = 20 m
t = 4.0 s

18. Example 3
A steel ball rolls with a constant velocity across a tabletop m high. It rolls off the table and hits the ground m from the edge of the table. How fast was the ball rolling just as it left the table?
dx = m
vx = ?
tx = ?
vi = 0 m/s
g = -9.8 m/s2
dy = m
ty = ?
vf = ?

19. Example 3 (continued) Horizontal Vertical G: U: E: S: t = 0.44 s
A steel ball rolls with a constant velocity across a tabletop m high. It rolls off the table and hits the ground m from the edge of the table. How fast was the ball rolling just as it left the table?
Horizontal Vertical
dy = m
vi = 0 m/s
a = -9.8 m/s2
dx = m G: U: E: S:
tx = ?
vx = ?
= 0.44 s
ty = ?
vf = ?
d = vit + ½ at2
d = ½ at2
t =
v = d/t 2d a
vx = (0.352 m) / (0.44 s)
t =
2 (0.950m)
(-9.8m/s2)
vx = 0.8 m/s
t = 0.44 s

20. Example 4
Divers at Acapulco dive horizontally from a cliff that is 65 meters high. If the rocks below the cliff protrude 27 meters beyond the edge of the cliff, what is the minimum horizontal velocity needed to safely clear the rocks below?
dx = 27 m
vx = ?
tx = ?
vi = 0 m/s
g = -9.8 m/s2
dy = 65 m
ty = ?
vf = ?

21. Example 4 (continued) Horizontal Vertical G: U: E: S:
Divers at Acapulco dive horizontally from a cliff that is 65 meters high. If the rocks below the cliff protrude 27 meters beyond the edge of the cliff, what is the minimum horizontal velocity needed to safely clear the rocks below?
Horizontal Vertical
dy = 65 m
vi = 0 m/s
a = -9.8 m/s2
dx = 27 m G: U: E: S:
tx = ?
vx = ?
= 3.64 s
ty = ?
vf = ?
d = vit + ½ at2
d = ½ at2
t =
v = d/t 2d a
vx = (27 m) / (3.64 s)
t =
2 (65 m)
(-9.8m/s2)
vx = 7.43 m/s
t = 3.64 s

22. Summary Projectile Motion is calculated using independent equations.
The horizontal movement of a projectile resembles constant motion.
The vertical movement of a projectile resembles freefall.
Time is the same for both vertical and horizontal components of projectile motion.