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. The initial height of the second ball is
(Take `g = 10 m s^(-2))`

Options

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

Answer

B

Explanation

Maximum height of projection , `H = (u^2 sin^(2) theta)/(2g)`
`:. H = (10)^(2)xxsin^(2) 30^(@) /(2xx10)=5/4=1.25 m`
Time for attaining maximum height, `t = (u sin theta)/g`
`:. t = (10 xx sin 30^@)/10 = 0.5 s`
`:.` Distance of vertical fall in 0.5 s ,`S = 1/2 "gt"^2`
or `S=1/2 xx 10 (0.5)^(2) = 12.5 m`
`:.` Initial height of the second ball
= S + H = 12.5 + 12.5 = 2.50 m