A stone is allowed to fall from the top of a tower 300m height and at the same time another stone is projected vertically up from the ground with a velocity `100ms^(-1)`. Find when and where the two stones meet?

Explanation

Let the two stones meet at C. The tiem of travel of the stones are equal . Vertically projected body travels a distance of x, while the freely falling body travels a distance of `(h-x)`.
For the stone moving upwards
`u=100ms^(-1), a =-9.8ms^(-1), t=?, s=x`
From the equation `s=ut+1/2at^(2), x=100t-1/2xx9.8t^(2)`
`x=100t-4.9t^(2)` ....(1)
For freely falling body
`u=0,a=9.8ms^(-2)`
`t=? s=h -x=(300-x)`
From eq.s =`ut+1/2xx9.8t^(2)`
`(300-x)=0+1/2xx9.8t^(2) (300-x)=4.9t^(2)` ......(2)
Adding equation (1)and(2)
`300=100trArrt=3sec`
substitute this value in equation (10
`x=100xx3-4.9xx9=300-44.1=255.9m`
the two stones meet after 3 sec and at a height of `255.9` from the ground.