A stone is allowed to fall from the top of a tower 100m 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.

Answer

The two stone meet after 4 sec at a height of 20m from the ground.

Explanation

Let the two stones meet at a height of x m from the ground after t sec. from the start.
For the downward motion of the stone A. u=0, `g=-10ms^(-2), s=-(100-x)`
`s=ut+1/2gt^(2)`

`-(100-x)=0 xx t+1/2 xx (-10) xx t^(2)`
`-(100-x)=-5t^(2)`
`-(100-x)=-5t^(2) .......(1)`
`u=25ms^(-2), g=-10ms^(-2), s=x`
`s=ut+1/2gt^(2)`
`x=25 xx t+1/2 xx (-10)t^2`
`x=25t-5t^(2) .......(2)`
From (1) and (2)
`-(100-x)=x-25t`
Substituting the value of t in (2)
`x=25 xx 4-5(4)^(2) therefore x=100-80`
x=20