A ball is droped from a high rise platform `t = 0` starting from rest. After `6 s` another ball is thrown downwards from the same platform with a speed `v`. The two balls meet at `t = 18 s`. What is the value of `v` ?

Options

A. 75 m/s
B. 55m/s
C. 40 m/s
D. 60 m/s

Answer

A

Explanation

Let two balls meet at depth h from platform
so `h=(1)/(2)g(18)^(2)=v(12)+(1)/(2)g(12)^(2)`
`implies v=75 ms^(-1)`