A stone is allowed to fall from the top of a tower $100m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25m{s}^{-1}$.Calculate where the two stones will meet.(Take $g=10m/s^2$).

The correct option is B 20 m

For stone moving downward,acceleration due to gravity ,$g=10m{s}^{-2}$.

Initial velocity (u)$=$o;distance, s$=$100-x;time, t$=$?

$s=ut+\frac{1}{2}g{t}^2$, or (100-x)=$0\times t+\frac{1}{2}\times 10t^2$

$\Rightarrow 100-x=5t^2$ .....(1)

For the stone moving vertically upwards,

For the stone moving vertically upwards,initial velocity, $u=25m{s}^{-1}$;time (t)=?;

acceleration due to gravity $g=-10m{s}^{-1}$

[In upward direction ,g is taken -ve]

Distance, $s=x$

we know $s-ut+\frac{1}{2}g{t}^2$ or

$x=25\times t+\frac{1}{2}\times (-10t^2)$

$\Rightarrow x=25t-5t^2$ .....(2)

Substituting the value of x from (2) in (1) we get,

$100-(25t-5t^2)=5t^2$

$100-25t+5t^2=5t^2$

$25t=100$ or $t=4s$

Put the value of tin (1) $\Rightarrow 100-x=5(4{)}^2$

$\Rightarrow 100-x=80$ or x$=$20 m

$\therefore$ The stones will meet at a height of 20m