Numerical based on gravitation

Given, H = 100 m
Let the stone thrown up be stone 1
v₁ = 25 m/s
a₁ = - g = - 10 m/s^₂ (as it is moving upward)

Let the stone released from H be stone 2
v₂ = 0
a₂ = g = 10 m/s²  (taking g = 10 m/s² approximately)

When the two stones meet then,

S₁+S₂ = H

i.e. (v₁t + 1/2 a₁ t² ) + (v₂ t + 1/2 a₂ t² ) = H

On substituing the values we get,

          (25xt - (1/2)10xt² ) + (0xt + (1/2)10xt² ) = 100

On solving we get,

            25xt = 100
i.e.             t = 4 seconds.

Thus the two stones will meet after 4 seconds.

Hope this helps.
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