Solve this question fast plz

Let the distance travelled by the stone which is dropped from the top upto the instant when the two stones meet be x 

Initial velocity u = 0

Acceleration a = g = 9.8 m s^-2

Using the second equation of motion
$$\begin{gathered} s = ut +\frac{1}{2}a{t}^2 \\ x = 0\times t+\frac{1}{2}\times 9.8\times t^2 \\ x =4.9t^2 ...\left(i\right) \end{gathered}$$

The distance travelled by the stone which is projected vertically upwards from the ground up to the instant when the two stones meet would be equal to 100 - x

Initial velocity = 25 m s^-1

Acceleration a = -g = -9.8 m s^-2

Using the second equation of motion
$$\begin{gathered} s = ut +\frac{1}{2}a{t}^2 \\ 100-x = 25\times t+\frac{1}{2}\times \left(-9.8\right)\times t^2 \\ 100-x = 25t-4.9t^2 \\ x =4.9t^2-25t +100 ...\left(ii\right) \end{gathered}$$

Equating x from (i) and (ii) we get

4.9t² = 4.9t²  - 25t + 100

25t = 100

t = 4s

x = 4.9t²

x = 4.9 X 4²

x = 78.4 m

100 - x = 100 - 78.4 = 21.6 m

The stones meet after a time of 4 seconds at a height of 21.6 metres from the ground.

Regards