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

Answer

c

Explanation

Let a ball takes time t to meet the another ball, the distance covered by ball is S.(t=18s)
`S=(1)/(2)gt^(2)=(1)/(2)g(18)^(2)`
Again a ball is thrown with velocity v and meet with previous ball, the displacement covered by ball will be same
`S=vt+(1)/(2)gt^(2)`
`S=v(t-6)+(1)/(2)g(t-6)^(2)`
Substituting from i
`(1)/(2)g(18)^(2)=v(18-6)+(1)/(2)g(18-6)^(2)`
`(g)/(2)(18)^(2)-(g)/(2)(12)^(2)=vxx12`
`v=(gxx30xx6)/(2xx12)`
`v=75m//s`