1. Topic 9.2 Space
9.2.2 Projectile Motion

2. Objectives
State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field.
Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance.
Describe qualitatively the effect of air resistance on the trajectory of a projectile.
Solve problems on projectile motion.

3. Projectile Motion

4. Components of Motion
When a body is in free motion, (moving through the air without any forces apart from gravity and air resistance), it is called a projectile.
Normally air resistance is ignored so the only force acting on the object is the force due to gravity
This is a uniform force acting downwards

5. Therefore if the motion of the projectile is resolved into the vertical and horizontal components
The horizontal component will be unaffected as there are no forces acting on it
The vertical component will be accelerated downwards by the force due to gravity

6. These two components can be considered as independent factors in the motion of a projectile in a uniform field
In the absence of air resistance the path taken by any projectile is parabolic

7. Solving Problems
In solving problems it is necessary to consider the 2 components independently

8. Comparison of Horizontal and Vertical Projectile Motion

9. Therefore for horizontal motion it is necessary to use the equation
speed = distance time
Where speed is the horizontal component of the velocity

10. Therefore for vertical motion it is necessary to use the kinematic equations for uniform acceleration

11. Example
A ball is kicked at an angle of 40.0o with a velocity of 10.0 ms-1. Taking g = 9.81 m.s–2. How far does it travel horizontally?
40o
10 m.s-1

12. To be able to calculate the horizontal distance we need to know the horizontal speed, and the time.
The horizontal distance is easy to calculate by resolving the velocity
10.0 sin 40.0o
40.0o
10.0 m.s-1
10.0 cos 40.0o

13. However, to calculate the time we will need to use the vertical component and the kinematics equations
s = ?
u = 10.0 sin 40.0o ms-1
v = ?
a = m.s-2 (Up is positive, therefore acceleration here is negative)
t = ?
We only have 2 of the values when we need three to find any other

14. However, if we ignore air resistance, then the final vertical component of the velocity will be equal and opposite of the initial component
i.e. v = sin 40.0o m.s-1
Looking at the equations for uniform acceleration, we need an equation that links u, v, a and t.

15. v = u + at
Rearranging to make t the subject
t = v – u a
Substitute in
t = sin 40.0o – 10.0 sin 40.0o
-9.81
t = 1.31 seconds

16. Now returning to the horizontal components
Using speed = distance time
Rearranging distance = speed x time
Distance = 10.0 cos 40.0o x 1.31
Distance = = 10.0 metres

17. Example
A golf ball is hit horizontally at 25.0 m.s-1 from the top of a 68.0 m cliff. Sketch the trajectory of the ball. How long does it take the ball to land? How far from the base of the cliff does the ball strike the ground? Find the impact velocity of the ball. [3.72 s, 93.1 m, ms-1, 56˚ with the horizontal]

18. Example
A golf ball is hit into the air at a velocity of 35.4 m.s-1 at an angle of 39.0 º to the horizontal. Make a sketch of the ball’s motion and find the:
Total time in the air. [4.55 s]
Time when maximum height is reached. [2.27 s]
Maximum height. [25.3 m]
Range (Horizontal distance) [125 m]

19. Example 1

20. Example 1 (cont.)

21. Example 1 (cont.)

22. Markscheme

23. Markscheme