A stone in dropped from a height 300m and at the same time another stone is projected vertically upwards with a velocity 100m/sec. Find when and where the two stones meet.

Let the two stones will meet at a height (x) m from the ground after a time t sec.

For stone –I
$\begin{array}{l}\mathrm{u}=0\\ \mathrm{a}=\mathrm{g}\\ \mathrm{s}=\mathrm{h}-\mathrm{x}\\ {\mathrm{t}}_1=\mathrm{tsec}\end{array}$
From $\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$
$\begin{array}{l}\Rightarrow \mathrm{h}-\mathrm{x}=0\left(\mathrm{t}\right)+\frac{1}{2}{\mathrm{gt}}^2\\ \mathrm{x}=\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2----1\end{array}$
For stone -II
$\begin{array}{l}\mathrm{u}=100\mathrm{m}/\sec \mathrm{a}=-\mathrm{g}\\ \mathrm{s}=\mathrm{x}, {\mathrm{t}}_2=\mathrm{tsec}\end{array}$
apply $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$
$\mathrm{x}=\left(100\right)\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2-----\left(2\right)$
From (1) &(2) $\Rightarrow 100\mathrm{t}=\mathrm{h};100\mathrm{t}=300$
$\mathrm{t}=3\sec$
From eq (2),

 $\mathrm{x}=100\times 3-\frac{1}{2}\times 9.8\times (3{)}^2$
$\mathrm{x}=300-44.1;\mathrm{x}=255.9\mathrm{m}$
The two stones will meet at a height 255.9 m from the ground after time 3 sec.