The velocity of a particle when at its greatest height is sqrt((2)/(5)) of its velocity when at half of its greatest height find the angle of projection

The velocity of a particle when at its greatest height is sqrt((2)/(5)

1.	The velocity of a particle when at its greatest height is sqrt((2)/(5)) of its velocity when at half of its greatest height find the angle of projection
Answer» Solution :Step - 1: we know that, velocity of a projectile at half of maximum height `= usqrt((1+ cos^(2) THETA)/(2))` Step 2 : given that `u cos theta = sqrt((2)/(5)) XX u sqrt((1+cos^(2) theta)/(2))` SQUARING on both sides `u^(2) cos^(2) theta = (2)/(5) u^(2) ((1+cos^(2) theta)/(2))` `10 cos^(2) theta = 2+ 2 cos^(2) theta` `RARR 8 cos^(2) theta = 2 rArr cos^(2) theta = (1)/(1) rArr theta = 60^(@)`