A stone is dropped from a height 300 m and at the same time another stone is projected vertically upwards with a velocity of `100m//sec`. Find when and where the two stones meet.

Explanation

Let the two stones meet at C.The time 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=100 `ms^(-1)1, a=-9.8 ms^(-1),t=?,s=x`
From the equation `s=ut+(1)/(2)at^(2),x=100t-(1)/(2)xx9.8t^(2)`
x=100t-`4.9t^(2)` ...........(1)
For freely falling body
u=0,a=9.8 `ms^(-2)`
t=?s=h-x=(300-x)
From eq.s=ut +`(1)/(2)xx9.8t^(2)`
`(300-x)=0+(1)/(2)xx9.8t^(2)" "(300-x)=4.9t^(2)`...........(2)
Adding equation (1)and (2)
300=100t` implies `t=3 sec
Substitute this value in equation (1)
`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.