1. 6.2 General projectile motion
Human cannonball
Projectile trajectory
Range and angle of projection
Time of flight and maximum height
Check-point 3 1 2 3
Book 2 Section 6.2 General projectile motion

2. Human cannonball
A man fired from a cannon lands precisely on the safety net.
How does he do it? How would you describe his trajectory?
He is projected at an angle.
His trajectory is a parabola.

3. 1 Projectile trajectory
Consider an object launched at an angle of projection  with initial velocity u :
Assume: No air resistance

4. 1 Projectile trajectory
Horizontal direction: uniform motion
sx = uxt
(1)
Since ux = u cos ,
 t = sx
u cos 
(2)
Vertical direction: free fall motion
sy = uyt at 2 1 2
(3)

5. 1 Projectile trajectory
Taking the upward direction as +ve, a = –g.
 sy = (u sin )t – gt 2 1 2
(4)
Substitute (2) into (4),
sy = (tan )sx – sx2 g
2u 2 cos2
… (5)
Equation of trajectory

6. 1 Projectile trajectory
Trajectory of an object: a parabola (when air resistance is negligible)
It has:
a symmetric path,
same time of upward & downward flights,
same speed for upward and downward motions at the same height

7. 1 Projectile trajectory
Trajectory travelled by projectile with air resistance has:
asymmetric path,
maximum height and horizontal distance much reduced
Golf
Example 5

8. A golf ball is projected at 15 to the horizontal with u = 30 m s–1
Example 5
Golf
A golf ball is projected at 15 to the horizontal with u = 30 m s–1
Horizontal distance s travelled = 45 m
Take g = 10 m s–2
Assume: Air resistance negligible

9. The trajectory is symmetrical.
Example 5
Golf
(a) Horizontal distance travelled by the ball as it reaches the max. height = ?
The trajectory is symmetrical.
 Horizontal distance travelled
=  45 1 2
= 22.5 m

10. The height of the ball is 1.19 m.
Example 5
Golf
(b) When horizontal distance travelled = 40 m, the height of the ball from the ground = ?
sy = (tan )sx – sx2 g
2u 2 cos2
= tan 15  40 –  402 10
2(302) cos2 15
= 1.19 m
The height of the ball is 1.19 m.

11. 2 Range and angle of projection
Range (sx): the horizontal distance travelled by a projectile
Suppose: launch and land at same level
Put sy = 0 into the equation of trajectory:
0 = (tan )sx – sx2 g
2u 2 cos2
 sx =
2u 2 sin  cos  g
u 2 sin 2 g =

12. 2 Range and angle of projection
Range (sx) of a trajectory :
sx =
2u 2 sin  cos  g
u 2 sin 2 g =
It depends on u and  :

13. 2 Range and angle of projection
For a given u :
From 0 to 45,
  
range 
From 45 to 90,
  
range 
u 2 g
At 45,
range =
(maximum)
Simulation
6.2 Range and angle of projectile

14. 2 Range and angle of projection
Example 6
Throwing darts at an angle

15. A dart is thrown towards a dartboard 2.4 m away.
Example 6
Throwing darts at an angle
A dart is thrown towards a dartboard 2.4 m away.
Height of dartboard = 0.46 m
The dart is thrown at the same level as the bullseye (centre of the dartboard ).

16. Example 6
Throwing darts at an angle
Can a dart reach a dartboard (2.4 m away) with the following velocities?
(a) 4 m s–1
(b) 10 m s–1
Use the graph to find out the angle(s) that the dart can hit the bullseye (at the same level as the point of projection).

17. Assume: Air resistance negligible
Example 6
Throwing darts at an angle
(a) Take g = 10 m s–2
Assume: Air resistance negligible
u 2 g
Max. range = = 42 10
= 1.6 m (< 2.4 m)
 The dart cannot reach the dartboard.

18.  The dart can hit the bullseye with a suitable .
Example 6
Throwing darts at an angle
(b)
Max. range
u 2 g = =
102 10
= 10 m (> 2.4 m)
 The dart can hit the bullseye with a suitable .

19.  The dart should be projected at 7 or 83 to the horizontal.
Example 6
Throwing darts at an angle
u 2 sin 2 g =
Range
102 sin 2 10
2.4 =
0.24
sin 2 = 0.24 7 83
 The dart should be projected at 7 or 83 to the horizontal.

20. 3 Time of flight and maximum height
Time of flight (t0):
The time interval during which the projectile is moving in the air
t0 =
2u sin  g
Maximum height H of projectile:
H =
u 2 sin2  2g
Example 7
Finding the maximum height of a golf ball

21. Take g = 10 m s–2. Neglect air resistance.
Example 7
Finding the maximum height of a golf ball
(a) A golf ball is projected at 15 to the horizontal with v = 30 m s–1.
Max. height of the ball = ?
Take g = 10 m s–2. Neglect air resistance.
u 2 sin2  2g
Max. height = =
(302) sin2 15
2 (10)
= 3.01 m

22. (b) Time of flight of the ball = ?
Example 7
Finding the maximum height of a golf ball
(b) Time of flight of the ball = ?
2u sin  g
Time of flight = =
2 (30) sin 15 10
= 1.55 s

23. 3 Time of flight and maximum height
Example 8
Shot-put

24. An athlete throws a 4-kg metal ball (shot) from a height of 1.8 m.
Example 8
Shot-put
An athlete throws a 4-kg metal ball (shot) from a height of 1.8 m.
Initial velocity = 10 m s–1 Angle of projection = 40
Take g = 10 m s–2. Neglect air resistance.

25. (a) Max. height of the shot = ?
Example 8
Shot-put
(a) Max. height of the shot = ?
Max. height =
u 2 sin2  2g =
(102) sin2 40
2(10)
= 3.87 m

26. (b) How far has the shot travelled upon touching the ground?
Example 8
Shot-put
(b) How far has the shot travelled upon touching the ground?
By sy = (tan )sx – sx2, g
2u 2 cos2
–1.8 = (tan 40)sx – sx2 10
2(102) cos2 40
0.0852sx2 – 0.839sx – 1.8 = 0

27. Solve the quadratic equation by sx:
Example 8
Shot-put
0.0852sx2 – 0.839sx – 1.8 = 0
Solve the quadratic equation by sx:
sx =
= 11.7 m or –1.81 m (rejected)
 The shot will go 11.7 m upon touching the ground.

28. (c) Time of flight of the shot = ?
Example 8
Shot-put
(c) Time of flight of the shot = ?
By sy = (u sin )t – gt 2 1 2
–1.8 = 10 (sin 40)t – (10)t 2 1 2
5t 2 – 6.43t – 1.8 = 0
t = 1.52 s or –0.24 s (rejected)
 The time of flight is 1.52 s.

29. Find the speed of the baseball when it flies off.
Check-point 3 – Q1
A batter hits a baseball at  = 60 to the horizontal. The baseball lands at the spectator seats.
Find the speed of the baseball when it flies off.

30. By , sy = (tan )sx – sx2 4 = (tan 60)(80) – (80)2 u = 30.8 m s–1
Check-point 3 – Q1
By ,
sy = (tan )sx – sx2
g 2u 2 cos2 
4 = (tan 60)(80) – (80)2 10
2u 2 cos2 60
u = 30.8 m s–1

31. An athlete takes off with 9 m s–1 at an elevation of 30.
Check-point 3 – Q2
An athlete takes off with 9 m s–1 at an elevation of 30.
Range of the jump = ?
u 2 sin 2 g
Range = =
92 sin 2(30) 10
= 7.01 m

32. The End