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))

A ball is projected from the ground at a speed of 10 ms^(-1)making an

1.	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))
Answer» 6.25 m 2.5 m 3.75 m 5 m Solution :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