Question

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. `74 m//s`
B. `64 m//s`
C. `84 m//s`
D. `94 m//s`

Answer 1

Correct Answer - A
For first ball, initial velocity `u = 0`
`s_1 = (1)/(2) g t_1^2 = (1)/(2) xx g xx (18)^2`
For second ball, initial velcity `u - v`
`s_2 = vt_2 + (1)/(2) g t^2`
`t_2 = 18 - 6 = 12 s`
`s_2 = v xx 12 + (1)/(2) g (12)^2`
Here, `s_1 = s_2`
`(1)/(2)g(18)^2 = 12 v + (1)/(2) g xx (12)^2`
`v = 74 m//s`.