Question

A rock is dropped from a height 'h'.Simultaneously,another rock is thrown up from the ground which creates a height'4h'.What would be the time taken to two rocks to cross each other?

Answer 1

Let the upward initial velocity of the second rock be `u_2`.

The maximum height that the 2nd rock can attain is `4h`

So we can write `0^2=u_2^2-4gh`, where g is the acceleration due to gravity.

Hence `u_2=sqrt(4gh)`

The initial velocity of the 1st rock dropped from height `h` is `u_1=0`.

Movement of the rocks is initiated at same moment, Let the rocks reach at same height after `t` s of their start. Here total distance traversed by the two rocks in `t` s should be `h`

So `(u_1xxt+1/2 g t^2)+(u_2xxt-1/2g t^2)=h`

`=>0xxt+cancel(1/2 g t^2)+sqrt(4gh)xxt-cancel(1/2g t^2)=h`

`=>t=h/sqrt(4gh)=1/2sqrt(h/g)`