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

Height h=300 m,
Initial velocity `U_0=100m//s`
Let the two stones will meet at a height `X` above the ground.
For 1st stone`h-x=1/2g t^2`…………(1)
For 2nd stone `x=u_0t-1/2g t ^2`
`implies1/2g t ^2=u_0t-x......(2)`

Since t is same for the two stones
From equations 1& 2.
`h-x=u_0t-x`
`impliesu_0t=hortimet=h/u_0=300/100=3 sec.`
`therefore ` The two stones will meet 3 seconds after the 1st stone is dropped or 2nd stone is thrown up.