From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25m/s. Find when and where the two balls meet.

for the ball dropped downwards

s=ut+1/2gt^2

s=1/2gt^2

s=4.9*t^2

 

For the ball thrown upwards

= 100-s=ut-1/2gt^2

100-s=25*t-4.9t^2

therefore

100-s=25t-49.t^2

Now add both the equations 

 100 =  25 t

So t = 4 seconds

 

S = 4.9 * 16 m