Question

A Ball Released from at rest time t = 0 hits the ground. If rebounded inelastically with a velocity 5ms^-1 and reaches the top at t = 1.5s . What is the net displacement of the ball from its initial position after 1.5s ?

Answer 1

› Rebound velocity of the ball, v = 5m/s

Now, v = u+gt
5 = 0 + 10t

.°. t = 5/10 = 0.5

(This is the time taken in rebound of the object)

So, time taken to reach the ground

= 1.5s - 0.5s = 1s

Again,

S= ut+1/2gt²

So, initial height = 0 + 1/2×10×1² = 5m

For the distance after rebounding of object

v²= u² + 2gh

So, h = 5²/2×10 = 1.25m

Displacement = 5-1.25 = 3.75m