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

Explanation

S = ut + `(1)/(2) "gt"^(2)`
= 0 `xx t + (1)/(2) xx 10 xx t^(2)`
`S_(1) = 5t^(2)`
`S_(2) = ut + (1)/(2) "gt"^(2)`
= 25 t + `(1)/(2)(-10)t^(2)`
`s_(2) = 25t - 5t^(2)`
`S_(1) + S_(2) = 100`m
`5t^(2) + (5t - 5t^(2)) = 100` M
t ` = (100)/(25) ` = 4 sec
`S_(1) = 5t^(2)`
= `5(4)^(2) = 80`M