Question

a stone is droped from a height of 400m high tower.at the same time another stone is projected vertically upward from the ground with a speed of 50m/s the stone will cross each other after a time . what is the time
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Answer 1

The answer to this problem will be that value of time when height of the stone dropped from ground will be equal to the height of projected stone.

Expression for distance of dropped stone from the ground will be

$s=ut+\frac{at^2}{2}\\s=0+\frac{gt^2}{2}$

$s=\frac{gt^2}{2}$

This is displacement from the top, so its height from ground will be

h=400-s         -----(1)

Now expression for height of stone projected=

$s=ut+\frac{at^2}{2}\\s=0+\frac{gt^2}{2}$

$s=50t-\frac{gt^2}{2}$ -------(2)

Equate (1) and (2):

$400- (gt^2/2)= 50t - (gt^2/2)\\400=50t\\t=8 sec$

Hence they meet after 8 seconds.