From the top of a tower 100 m in height, a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 `"m.s"^(-1)`. Find when and where the two balls meet. Take g = 9.8 `"m.s"^(-2)`.

Explanation

AB = height of the tower = 100 m . Let the two balls meet after a time t at the point C, such that AC = `h_(1) ,` BC= `h_(2)` and `h_(1)+h_(2)` m.
For the first ball `h_(1) = (1)/(2) "gt"^(2)`
For the second ball `h_(2) = ut - (1)/(2) "gt"^(2)`
By adding we get
`h_(1)+h_(2)` = ut
or , 100 = 25 t or, t = 4 s
So, the two balls meet after 4 s .
The height of the point C , where the two balls meet from the ground is
`h_(2) = ut - (1)/(2)"gt"^(2)`
`= 25 xx 4 - (1)/(2)xx 9.8 xx4^(2)` = 100 - 78.4 = 21.6 m