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

Answer

A

Explanation

Let the two balls meet after t s distance x from the platform.
For the first ball u = 0 , t = 18 s , g = `10 m//s^2`
Using `h = "ut" +1/2 "gt"^2`
`:. x = 1/2 xx 10 xx 18^(2) " "...(i)`
For the second ball u = v, t = 12 s , ` g = 10 m/s^(2)`
Using `h = ut +1/2 "gt"^(2)`
`:. x = v xx12 +1/2 xx10 xx 12^(2) " "...(ii)`
From equations (i) and (ii) , we get
`1/2 xx10 xx 18^(2) = 12 v +1/2 xx 10 xx (12)^(2)`
or `12 v =1/2 xx 10 xx [ (18)^(2) - (12)^2]`
or `v = (1xx10 xx30 xx 6)/(2xx12) = 75 m//s`