Question

A ball is dropped from height of 20m at the same instant and the ball is thrown up from the ground with speed 20m/s when and where will ball meet​

Answer 1

After 1 second, the dropped ball has fallen:

s = ut + 1/2 gt^2

s = 0 + 1/2 (-10) 1^2

s = - 5 m from starting point

After 1 second, the thrown ball has risen:

s = ut + 1/2 gt^2

s = 20 (1) + 1/2 (-10) 1^2

s = 20 -5 m

s = + 15 m from starting point

The two balls meet one second after release and meet at a point 15 meters above ground.