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

Answer

A

Explanation

Clearly distance moved by `1^(st)` ball in 18s= distance moved by `2^(nd)` ball in 12s.
Now, distance moved in 18s by `1^(st)` ball `=(1)/(2) xx 10 xx 18^(2) = 90 xx 18= 1620m`
Distance moved in 12s by `2^(nd)` ball `=u t + (1)/(2) g t^(2)`
`:. 1620 = 12v + 5 xx 144 rArr v= 75ms^(-1)`