From the top of the tower of height `400 m`, a ball is dropped by a man, simultaneously from the base of the tower, another ball is thrown up with a velocity `50 m//s`. At what distance will they meet from the base of the tower ?

Options

A. 100 m
B. 320 m
C. 80 m
D. 240 m

Answer

C

Explanation

`400 - s = (1)/(2) g t^2`…(i)
and `s = 50 t - (1)/(2) g t^2`…(ii)
Adding, `50 t = 400`
or `t = 8 sec`
`s = 50 xx 8 - (1)/(2) xx 10 xx 64 = 80 m`.