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

1.	The velocity of a projectile 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: 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))xxusqrt((1+cos^(2)theta)/(2))` SQUARING on both SIDES `u^(2)cos^(2)theta = (2)/(5)u^(2)((1+cos^(2)theta)/(2))` `10costheta = 2 + cos theta` `RARR 8cos^(2)theta = 2 rArr cos^(2)theta=(1)/(4)rArr theta=60^(@)`