1. Projectile Motion

2. Projectiles
Projectile – an object that is “projected” or shot into the air; once air born it has no means of altering its path. As such a projectile is an object upon which the only force acting on it is gravity.
Types of projectiles - Cannon balls, spears, arrows are all considered to be projectiles.
Click on the picture to go to the PhET site and open the Projectile Motion SIM. Play with the SIM for a while.

3. Examples of Projectiles
Non-Projectiles
Thrown rock
Rocket
Fired bullet
Missile
Dropped bomb
Powered plane
Arrow
Bird (unless it is dead!)

4. Path of Projectile
Inertia: an object is in constant motion unless acted on by another force.

5. Path of Projectile
Path with no gravity!
With air friction
If friction is ignored (i.e. no air, or projectile shot on the moon) then:
Three Big Rules for Projectile Motion:
1.    The only force acting on a projectile is gravity!
2.    Therefore horizontal velocity will remain constant.
3. The vertical velocity is affected by the acceleration of gravity.

6. Path of Projectile
Horizontal Motion
Vertical Motion
Accele-ration No
Yes, “a” is downward at 9.8 m/s2
Velocity
Constant
Changing by 9.8 m/s every second
Since we have objects going in two directions; 1) along the x axis and 2) along the y axis we have to divide information as horizontal motion (ax , vx) and vertical motion (ay , vy)

7. Simple Projectile
To fully analyze projectiles you need to understand vectors. In this course we are going to examine one type of simple projectiles that requires very little knowledge of vectors. The more complicated projectile motion will be covered in Physics 12
Our projectiles will be projected horizontally off a cliff:
v = 20 m/s
h= 34 m

8. The motion of this horizontally projected cannon ball
The motion of this horizontally projected cannon ball. It has an initial horizontal velocity, vx = 20 m/s.
1) What do you notice about the horizontal velocity of the projectile all through out it’s motion?
It stays the same 
2) Now examine its vertical velocity, vy. What do we notice about the vertical velocity?
It decreases 9.8 m/s every second.

9. After watching the cannon ball’s projections off the cliff, the following data can be surmised.
Time
Horizontal Velocity
Vertical Velocity
0 s
20 m/s, right
1 s
9.8 m/s, down
2 s
19.6 m/s, down
3 s
29.4 m/s, down
4 s
39.2 m/s, down
5 s
49.0 m/s, down
NOTICE THE INITIAL VERTICAL VELOCITY….. VERY IMPORTANT

10. In fact the vx remains constant while the vy increase (downwards) due to the acceleration of gravity. These two facts will allow us to determine the range of any horizontally shot projectile.
Since the horizontal velocity remains constant (in the absence of air resistance) then the RANGE (distance from start to final impact) of a projectile, dx, can easily be calculated by:
dx = (vx )( t)

11. If we are given the horizontal velocity and we know the height of the cliff then we can calculate the range of the projectile, only if we know the time it takes for the projectile to land on the ground
Out problem is how do we solve for time?
We must first determine how long it takes for the ball to drop.
Remember the only acceleration on the projectile is the acceleration due to the gravitational force, g = 9.80 m/s2.
Therefore a dropped ball will “fall” at the same rate as a horizontally projected ball!
v = 20m/s
h = 34 m
dx = ( vx )(t)

12. Take two cannon balls, drop one from the top of the cliff as another one is shot horizontally off the same cliff.
Note how the “dropped” ball and the projected ball travel the same vertical distance in the diagram shown below:
The time for both of the balls to fall to the bottom of the cliff is exactly the same!
Of course the projected ball will land down range, but they will both hit the ground at the same time!
Don’t believe me! Then check out this Mythbusters clip!
We have dealt with falling objects before and therefore should be able to determine the time for the fall of the vertical ball (and therefore the projected ball) from a simple kinematic equation.

13. Example #1 2 x dy t = a 2 x -34 t = -9.80 t = 2.63 s
A cannon is shot at 20 m/s off a 34 m high cliff determine the time it takes to hit the ground and the range.
We can use the equation dy = viyt + ½ at2
Since the initial velocity of the vertical drop is 0 m/s we can deduce the equation to  dy = ½ at2
It can be rewritten as: a
t =
2 x dy
-9.80
t =
2 x -34
t = s

14. dx = (vx)( t) dx = 20 m/s x 2.63 s dx = 52.6 m = 53 m
Since we now know both the horizontal velocity and the time we can determine the Range of this projectile:
dx = (vx)( t)
dx = 20 m/s x s
dx = m
= 53 m

15. Best Practices to solve Questions
1) Draw out the situation
2) Divide you page into horizontal information and vertical information
3) Identify your known and unknown values
4) Solve

16. Example #2 Horizontal (x) Vertical (y) dix = ??? diy = -67 m
1. An arrow is shot horizontally off a 67 m high cliff with a velocity of 82 m/s. Determine the range of this projectile.
(Your answer should be 303 m)
Step 1 - Draw it out
Step 2 and 3 -
Divide info into Horizontal and Vertical. Then identify known and unknowns
Vix = 82 m/s
h = 67 m
Horizontal (x)
Vertical (y)
dix = ???
diy = -67 m
vix = 82 m/s
viy = 0 m/s
ax = 0 m/s2
ay = -9.8 m/s2
t = ??

17. Step 4 – Solve question
Part 2 –
dx = vix t
dx = (82m/s)(3.6977s)
dx = m
= 3.0 x 102 m
Part 1 – find the time
dy = (viy)(t) + ½ at2
-67 m = (0 m/s)(t) + ½ (-9.8 m/s2)t2
-67 m = 0 – 4.9 m/s2 (t2)
−67𝑚 −4.9 𝑚 𝑠 /𝑠 = t2
−67𝑚 −4.9 𝑚 𝑠 /𝑠 = 𝑡 2
t = ….s
t = 3.7 s

18. 2. A blue ball is shot horizontally of a cliff at 23 m/s
2. A blue ball is shot horizontally of a cliff at 23 m/s. It lands at a range of 132 m away from the base of the cliff. What is the height of the cliff? (ans. = 161 m)
3. A red ball is shot horizontally of a cliff and takes 4.5 s to hit the ground. It lands at a range of 132 m away from the base of the cliff.
What is the height of the cliff? (ans. = -99 m)
What was the horizontal velocity of the ball? (ans. = 29 m/s)
4. A green ball is projected horizontally at 45 m/s off a 27 m high cliff at the same time as a brown ball is dropped from the same cliff.
Which ball will hit the ground first? (ans. = at same time)
What is the vertical impact velocity of each ball with the ground? (ans. = -23 m/s)