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

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.

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

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)

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

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

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

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

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

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.

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 =

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  :

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

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

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 ).

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).

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.

 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 .

 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.

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

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

(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

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

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.

(a) Max. height of the shot = ? Example 8 Shot-put (a) Max. height of the shot = ? Max. height = + 1.8 u 2 sin2  2g = + 1.8 (102) sin2 40 2(10) = 3.87 m

(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

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.

(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.

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.

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

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

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