1.

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` ?A. 75 m/sB. 55m/sC. 40 m/sD. 60 m/s

Answer» Correct Answer - A
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)`


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