A man is standing on top of a building $100m$ high. He throws two balls vertically upwards, one at $t=0$ and other after a time interval (less than $2s$). The vertical gap between first and second ball is $15m$ at $t=2s$. The gap is found to remain constant. The velocities with which the balls were thrown are ( Take $g=10m{s}^{-2}$)

For first stone, 

Taking the vertical upwards motion of the first stone up to highest point

Here, $u=u_1, v=0$ (At highest point velocity is zero)

$a=-g, S=h_1$

As $v^2-u^2=2aS$

$\therefore {\left(0\right)}^2-{u_1}^2=2\left(-g\right)h_1$ or $h_1=\frac{{u_1}^2}{2g}$   … (i)

For second stone, 

Taking the vertical upwards motion of the second stone up to highest point

Here, $u=u_2, v=0, a=-g, S=h_2$

As $v^2-u^2=2aS$

$\therefore {\left(0\right)}^2-{u_2}^2=2\left(-g\right)h_2$ or $h_2=\frac{{u_2}^2}{2g}$   … (ii)

As per question

$h_1-h_2=15m, u_2=\frac{u_1}{2}$

Subtract (ii) from (i), we get; $h_1-h_2=\frac{{u_1}^2}{2g}-\frac{{u_2}^2}{2g}$

On substituting the given information, we get

$15=\frac{{u_1}^2}{2g}-\frac{{u_1}^2}{8g}=\frac{3{u_1}^2}{8g}$ or ${u_1}^2=\frac{15\times 8g}{3}=\frac{15\times 8\times 10}{3}=400$

or $u_1=20m{s}^{-1}$ and $u_2=\frac{u_1}{2}=10m{s}^{-1}$