A stone is dropped from the top of a tower 400 m high. At the same time another stone is thrown upwards from the ground with a velocity of 100 `"m.s"^(-2)`. When and where will they meet each other? [g = 9.8 `"m.s"^(-2)`]

Explanation

Let the two stones meet after a time t at a distance h from the top of the tower.
Considering the downward motion of the 1st stone,
`h = (1)/(2) "gt"^(2) = (1)/(2) xx9.8t^(2) " "cdots(1)`
Considering the upward motion of the 2nd stone
`400 -h = 100 t - (1)/(2)xx 9.8t^(2) " "cdots(2)` From equations (1) and (2) we get ,
`400 - (1)/(2) xx9.8t^(2) = 100t - (1)/(2)xx9.8t^(2)`
or,`" " 100t = 400 " ""or", t= 4s`
Hence from equation (1) we get ,
`h = (1)/(2) xx 9.8 xx(4)^(2) = (1)/(2) xx 9.8 xx16 = 78.4` m
Hence the two stones meet at 78.4 m below the top of the tower after 4 s.