Let the two stones meet after a time $t$.

When the stone dropped from the tower

Initial velocity,$ u = 0 \,m/s$

Let the displacement of the stone in time $t$ from the top of the tower be $s$.

Acceleration due to gravity, $g = 9.8\, ms^{-2}$

From the equation of motion,

$s=u t+\frac{1}{2} a t^{2}$

$s=0 \times t+\frac{1}{2} \times 9.8 \times t^{2}$

$\Rightarrow s=4.9 t^{2}$………….$(1)$

When the stone thrown upwards

Initial velocity, $u = 25\, ms^{-1}$

Let the displacement of the stone from the ground in time $t$ be $s^{\prime}$.

Acceleration due to gravity, $g = -9.8\, ms^{-2}$

Equation of motion,

$\quad s=u t+\frac{1}{2} a t^{2}$

$s^{\prime}=25 \times t-\frac{1}{2} \times 9.8 \times t^{2}$

$\Rightarrow s^{\prime}=25 t-4.9 t^{2}……………..(2)$

The combined displacement of both the stones at the meeting point is equal to the height of the tower $100\, m$.

$s^{\prime}+s=100$

$\Rightarrow 25 t-4.9 t^{2}+4.9 t^{2}=100$

$\Rightarrow t=\frac{100}{25} s=4 s$

In $4 s$,

The falling stone has covered a distance given by $(1)$ as $s=4.9 \times 4^{2}=78.4\, m$

Therefore, the stones will meet after $4 \,s$ at a height $(100 – 78.4) = 20.6 \,m$ from the ground.