Question

A stone is allowed to fall from the top of a tower 100 m
and at the same time another stone is projected vertically
upwards from the ground with a velocity of 25 m/s. Calculate
when and where the two stones will meet.​

Answer 1

Let the stones meet at point A after time t.

For upper stone  :

u′  = 0

x = 0 +  1/2 gt^2

x = 1/2 x 10 x t^2

⟹x = 5t^2  ............(1)

For lower stone :

u = 25  m/s

100 − x = ut − 1/2gt^2

100 − x = (25)t −  2 1 × 10 × t^2

⟹100 − x = 25t − 5t^2   ............(2)

Adding (1) and (2),  we get          

25t = 100              

⟹  t = 4 s

From (1),            

x = 5 × 4^2

⟹x=80 m  

Hence the stone meet at a height of   20 m above the ground after 4 seconds.