# Q.A ball is released from rest at a height "h" above the ground.The ball collides with the ground and bounces up at 75% of the impact speed the ball had with the ground.The collision with the...

Q.A ball is released from rest at a height "h" above the ground.The ball collides with the ground and bounces up at 75% of the impact speed the ball had with the ground.The collision with the ground is nearly instantaneously.A second ball is released above the first ball from the same height the instant the first ball loses contact with the ground.Calculate the time when the two balls collide.

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### 1 Answer

A ball is released from rest at a height "h" above the ground. The ball collides with the ground and bounces up at 75% of the impact speed the ball had with the ground. The collision with the ground is nearly instantaneously. A second ball is released above the first ball from the same height the instant the first ball loses contact with the ground.

The speed of the first ball when it strikes the ground is equal to v where v^2 - 0 = 2*g*h

=> v = `sqrt(2*g*h)`

The speed with which the ball bounces back up is `(3/4)*sqrt (2*g*h)`

If the balls collide t seconds after the first leaves the ground, `(3/4)*sqrt (2*g*h)*t - (1/2)*g*t^2 + (1/2)*g*t^2 = h`

=> `(3/4)*sqrt (2*g*h)*t = h`

=> `sqrt (2*g*h)*t = (4/3)*h`

=> `t = (4*h)/(3*sqrt (2*g*h))`

The two balls collide `(4*h)/(3*sqrt (2*g*h))` seconds after the second ball is left.