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» <html><body><p></p>Solution :If u and `alpha` be the <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> velocity of projection and anggle of projection respectively, then the maximum height attained `H_(m)=(u^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)sin^(2)alpha)/(2g)` and horizontal range `R=(2u^(2)sin alpha cos alpha)/g` <br/> <a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to the <a href="https://interviewquestions.tuteehub.com/tag/problem-25530" style="font-weight:bold;" target="_blank" title="Click to know more about PROBLEM">PROBLEM</a> we can write <br/> `(2u^(2)sin alpha cos alpha)/g=2sqrt(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)((u^(2)sin^(2)alpha)/(2g))impliestan alpha=(2/(sqrt(3)))impliesalpha=tan^(-1)(2/(sqrt(3)))`</body></html>