A stone is allowed to fall from the top of the tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of `25ms^ -1`. Calculate when and where the two stones will meet ?

Explanation

Let the two stones meet after a time t second at a distance of x from the ground.
Then distance covered by the stone allowed to fall from the top of the tower
`199 - x = u t + (1)/(2) g t^(2)`
`= 0 xx t + (1)/(2) g t^(2) = (1)/(2) g t^(2)` .....(1)
And the distance covered by the stone thrown from the ground
`x = u t - (1)/(2) g t^(2)`
`= 25 t - (1)/(2) g t^(2)` ......(2)
Adding (1) and (2), we get
`100 = 25t rArr t = 4` seconds.
And `x = 25 xx 4 - (1)/(2) xx 9.8 xx 4^(2)`
`= 100 - 78.4 = 21.6 m`.