Two projectiles are projected at angles `( theta)` and `((pi)/(2) - theta )` to the horizontal respectively with same speed `20 m//s`. One of them rises 10 m higher than the other. Find the angles of projection. (Take `g= 10 m//s^(2)`)

Explanation

Maximum height `H=(u^(2)sin^(2)theta)/(2g)rArrh_(1)=((20)^(2)sin^(2)theta)/(2g)=20sin^(2)theta & h_(2)=((20)^(2)sin^(2)(pi//2-theata))/(2g)=20 cos ^(2) theta`
`h_(2)-h_(1)=20 [cos ^(2)theta-sin^(2)theta]=10rArr20 cos 2 theta=10rArr cos 2 theta=(1)/(2)rArr(1)/(2)rArr2theta=60^(@)rArrtheta=30^(@)`
and `theta'=90^(@)-60^(@)`