A stone is thrown vertically upwards from the ground with a velocity 15 m/s. At the same instant a ball is dropped from a point directly above the stone from a height of 30 m. At what height from the ground will the stone and the ball meet and after how much time? (Use `g= 10 m//s^2` for ease of calculation).

Explanation

Let us assume that the stone and the ball meet after time `t_0` . The distances (not displacements) travelled by the stone and the ball in that time can be obtained from Eq. (3.8) as
`S_("stone") = 15t_0 -1/2 g t_0^2`
`S_("ball") =1/2 g t_0^2`
When they meet, `S_("stone") + S_("ball") = 30`
`15 t_0 -1/2 g t_0^2 + 1/2 g t_0^2 = 30`
`t_0 = 30//15 = 2 s`
`:. S_("stone") = 15 (2)-1/2(10) (2)^2 = 30 -20 =10 m`
Thus the stone and the ball meet at a height of 10 m.