(plz ans quickly very urgent) my question is

Let the stone which is dropped from the terrace be A and the stone thrown upwards be  B

Take the vertical downward motion as the positive direction of  acceleration of gravity.

Let the stones meet at a depth of x from the top of the building.

Then the height from the ground would be (100m - x)
For the stone A:
initial velocity u = 0m/s
acceleration a= 9.8m/s² 
Distance covered s = x
time taken = t

s= ut +(1/2)at²

x= 0+(1/2)x9.8 x t²     => x = 4.9t²

For stone B:
initial velocity u = 25m/s
acceleration a= -9.8m/s² 
Distance covered s = 100-x
time taken = t ( as they meet at the same time)

s= ut +(1/2)at²

100- x = 25 t + (1/2)  ( - 9.8)  t²    

=> 100- 4.9t² = 25 t + (1/2)  ( - 9.8)  t²    

Solving it we get,

t = 4sec

And x = 4.9t² .Substituting the value of t we get x = 78.4m

Hence the stones meet after 4 sec at a depth of 78.4 metres below the top of the building.