A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of `40ms^(-1)` from the bottom of the building. The two balls will meet after.

Options

A. 3 s
B. 2 s
C. 2.5 s
D. 5 s

Answer

C

Explanation

Let balls meet after t s.
Distance travelled by the ball coming down is
`s_1 = 1/2 "gt"^2`
Distance travelled by the ball thrown upward is
`s_(2) = 40 t -1/2 "gt"^2`
`:.s_(1)+s_(2)=100 m, :.1/2"gt"^2 +40t -1/2 "gt"^(2) = 100m`
or ` t = 100/400 = 2.5 s `