The horizontal range of a projectile is 2sqrt(3) times its maximum height. Find the angle of projection.

The horizontal range of a projectile is 2sqrt(3) times its maximum hei

1.	The horizontal range of a projectile is 2sqrt(3) times its maximum height. Find the angle of projection.
Answer» Solution :If u and `alpha` be the INITIAL velocity of projection and anggle of projection respectively, then the maximum height attained `H_(m)=(u^(2)sin^(2)alpha)/(2g)` and horizontal range `R=(2u^(2)sin alpha cos alpha)/g` ACCORDING to the PROBLEM we can write `(2u^(2)sin alpha cos alpha)/g=2sqrt(3)((u^(2)sin^(2)alpha)/(2g))impliestan alpha=(2/(sqrt(3)))impliesalpha=tan^(-1)(2/(sqrt(3)))`