A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is thrown vertically upwards from the ground wiht a velocity of 25m/s. Calculate when and where the two stones will meet. Support your answer with the help
Numerical based on gravitation
Given, H = 100 m
Let the stone thrown up be stone 1
v1 = 25 m/s
a1 = - g = - 10 m/s2 (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.