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

The correct option is B $2.5s$

Total height $=100m=S_1+S_2$

Using equation of motion,

For first ball

$S_1=ut+\frac{1}{2}g{t}^2$

where,

$S=$ Distance

$u=$ Initial velocity

$g=$ acceleration due to gravity

$t=$ time

$S_1=0-\frac{1}{2}g{t}^2$

For second ball

$S_2=40t+\frac{1}{2}g{t}^2$

$S_2=40t-S_1$ .......(i)

Putting the value of $S_1$ in equation (i)

$S_2+S_1=40t$

$100=40t$

$t=2.5sec$

Hence, two balls will meet after $2.5sec$