Question

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. For what value of h does the collision occur at the instant when the first ball is at its highest point?

Answer 1

`h= (v_(0)^(2) )/(g)`

Step-by-step explanation:

The height reached by the first ball as function of time can be given by the following equation:

`h_(1)(t)= v_(0)t - (g*t^(2) )/(2)`

The time required for the first ball to reach its highest point would be equal to its initial speed divided by the acceleration exerted by gravity. Therefore, maximum height can be found as follows:

`t=(v_(0))/(g) \\h_(1)((v_(0))/(g))= (v_(0))/(g)*v_(0) - (g*v_(0)^(2) )/(2g^(2))\\h_(1)((v_(0))/(g))= (v_(0)^(2))/(2g)`

Meanwhile, the distance that second ball falls during this time can be found by:

`h_(2)((v_(0))/(g))= (g*t^(2) )/(2)=(g*v_(0)^(2) )/(2g^(2)) = (v_(0)^(2))/(2g)`

The value h from which the second ball would have to be dropped in order for the collision to happen at the firs ball's highest points is found as follows:

`h = h_(1)+h_(2) = (v_(0)^(2) )/(2g)+(v_(0)^(2) )/(2g) \\h= (v_(0)^(2) )/(g)`