A ball is projected upwards from a point 4m above the ground with a speed of 4m/s. Find:
(a) The speed of the ball when it is 15m above its point of projection
(b) The speed with which the ball hits the ground.
1 Answer | Add Yours
The ball is projected upwards with a speed of 4 m/s from a point 4 m above the ground. There is an acceleration on the ball due to the gravitational force of attraction of the Earth equal to 9.8 m/s^2 acting downwards.
Let the speed of the ball when it is 15 m above its point of projection be v.
Use the formula v^2 - u^2 = 2*a*s, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance traveled.
v^2 - 4^2 = 2*15*(-9.8)
=> v^2 = -278
This is not possible. We can infer that the ball cannot reach a height of 15 m if it is projected at 4 m/s.
The maximum height that the ball can reach is 16/2*9.8 = 0.816 m.
The velocity of the ball when it hits the ground is:
v^2 - 16 = 2*4*9.8
=> v = 9.715 m/s
=> v = 8.85 m/s
We’ve answered 319,809 questions. We can answer yours, too.Ask a question