A ball is projected from the ground at a speed of `10 ms^(-1)` making an angle of `30^(@)` with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. Both the balls collide at the maximum height of first ball. The initial height of the second ball is (`g = 10ms^(-2)`)

Options

A. 6.25 m
B. 2.5 m
C. 3.75 m
D. 5 m

Answer

B

Explanation

`H =(u^(2) sin^(2) theta)/(2g) = ((10^(2) xx sin^(2) 30)/(2 xx 10)) = 1.25 m`
`t = (u sin theta)/g` [time to reach the top point]
`=(10 xx 1/2)/10 = 0.5` sec
Distance of vertical fall in 0.5 sec
`S = 1/2 "gt"^(2) =1/2 xx 10 xx (0.5)^(2) = 1.25 m`
Height of second ball = `1.25 +1.25 = 2.5 m`.