A ball `A` is projected with speed `20 m//s` at an angle of `30^(@)` with the horizontal from the ground. Another ball `B` is released simultaneously from a point on the vertical line along the maximum height of the projectile. Both the balls collide at the maximum height of the first ball. The initial height of ball `B` is

Options

A. `5 m`
B. `10 m`
C. `15 m`
D. `20 m`

Answer

B

Explanation

Time taken by `A` reach at highest point,
`t = (20 sin 30^(@))/(g) = 1 s`
`H_(max) = ((20)^(2) sin^(2) 30^(@))/(2g) = 5 m`
`h = (1)/(2) g t^(2) = (1)/(2) xx 10 xx (1)^(2) = 5 m`
Initial height of ball `B = h + H_(max)`
`= 10 m`