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

The correct option is B $2.5\text{s}$
Let the distance from top of the building where the balls meet be $x$,
(Taking downward as positive)
For first ball,
$\begin{array}{}x=\frac{1}{2}g{t}^2& \text{(1)}\end{array}$
For the second ball, $\begin{array}{}-(100-x)=-40t+\frac{1}{2}g{t}^2& \text{(1)}\end{array}$
Adding $\left(1\right)$ and $\left(2\right)$,
$40t=100$
$t=\frac{100}{40}=2.5\text{s}$

Using Relative velocity:
Initial velocity of stone 1 which is dropped from the top of a tower, $u_1=0$
Initial velocity of stone 2 which is thrown upward from the ground, $u_2=+40\text{m/s}$
(Take upward direction to be positive)
Hence,
$u_{rel}=u_{21}=u_2-u_1=40-0=40\text{m/s}$
$S_{rel}=100\text{m}$
$a_{rel}=g-g=0$
Using second equation of motion, we get
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Rightarrow 100=40\times t+0$
$\therefore t=2.5\text{s}$
Hence, the two stones will meet after 4 seconds.