Cambridge Maths Textbook 3U - Projectile Motion Question (1 Viewer)

[confusius]

Exercise 3G Q10

Gee Ming the golfer hits a ball from level ground with an initial speed of 50m/s and an initial angle of elevation of 45◦. The ball rebounds off an advertising hoarding 75 metres away. Take g = 10m/s 2.

(a) Show that the ball hits the hoarding after 3/2 √2 seconds
at a point 52·5 metres high.
(b) Show that the speed v of the ball when it strikes the hoarding is 5√58m/s at an angle of elevation α to the horizontal, where α = tan−1 (2/5)
(c) Assuming that the ball rebounds off the hoarding at an angle of elevation α with a speed of 20% of v, find how far from Gee Ming the ball lands.

I have already proven part a) and part b). But part c) really confuses me. I have tried many times but I have not solved it. How do you solve part c)? Do you work out the six equations of motions of the ball bouncing back? What values do you substitute in?
Please help!!!

 

[HoldingOn]

Basically yes we derive the equations for the motion of the ball once it has hit the advertising hoarding.
1. Set up the vector resolution (the right-angled triangle) for after the ball hits the sign with the information given in the question, i.e. V= 0.2(5√58). α = tan−1 (2/5).
2. Work out expressions for vertical velocity and vertical displacement (remember to include 52.5 as the constant as the ball hits the sign this high and thus is initially this high). Find the time when vertical displacement is 0, i.e. when the ball hits the ground.
3. Sub that value for T into the horizontal displacement which you have derived. Then subtract the distance you get from 75 as it is given that the sign is 75m away.