A stone is allowed to fall from the top of a tower 300m height and at the same time another stone is projected vertically up from the ground with a velocity 30m/s .Find when the two stones meet?

Let ‘t’ be the time after which the two stones meet
$\therefore$ For a vertically projected upwards body  $u=30m/s$

$\begin{array}{l}a=g=-9.8\mathrm{m}/{\mathrm{s}}^2\\ s=xt=t\end{array}$

$\therefore$ From Equation  $s=ut+\frac{1}{2}a{t}^2$
   $x=30t-\frac{1}{2}g{t}^2\mathrm{...........}\left(i\right)$
For a freely falling body
$\begin{array}{l}u=0\\ a=g=9.8\mathrm{m}/{\mathrm{s}}^2\\ s=y\\ t=t\\ \therefore \text{ From equation }s=ut+\frac{1}{2}g{t}^2\end{array}$

$\begin{array}{l}y=\frac{1}{2}g{t}^2\dots \dots \dots ..\left(2\right)\\ \text{ Given that }x+y=300\\ \Rightarrow 30t=300\\ \Rightarrow t=10\mathrm{s}\end{array}$