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

If the two balls meet at a distance ‘x’ from the bottom of the tower.

(100 - x) = 1/2gt²    ….(1)

and  h = ut + 1/2at²

so,  x = 25t – (1/2)gt² ….(2)

From eqn (1) & (2),we get,

100 = 25t

t = 4sec

From eqn (2)

x = 25 *4 – (1/2)(9.8)(4)²

x = 21.6m

Therefore, the height at which the two balls will meet = 21.6m and time, t=4sec.