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

Answer

B

Explanation

For 1 ball: `h=(1)/(2)"gt"^(2)=(1)/(2)xx10xx18^(2)=1620m`
For II ball : `(h=1620m, u=v, t=19-6=12s)`
`h=ut+(1)/(2)"gt"^(2)`
`1620=12v+(1)/(2)xx10xx12^(2)`
`1620=12v+720`
`12v=900rArr v=75m//s`