6.2 General projectile motion

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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