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

Answer

B

Explanation

The ball meet at height .h.
`h = 1/2 g t^2 = 1/2 g xx (18)^2`
`h = ut + 1/2 + 1/2 g t^2 = v (18-6) + 1/2 g (18-6)^2 `
Now , `1/2 xx g xx (18)^2 = v xx 12 + 1/2 xx g xx (12)^2`
`rArr v = (1/2 xx 10(18^2 - 12^2) )/(12) = (1/2 xx 10 xx 30 xx 6)/(12)`
`v = 75 m//s`