Question

the horizontal rays of a projectile is 2√3 times of ita maximum hight find angle of projection ? ​

Answer 1

Given that,

$\sf \: R = 2 \sqrt{3} H$

Range and Height of a projectile can be expressed as :

$\sf \: 4H = R \tan( \alpha)$

Therefore,

$\implies \sf \: 4H =2 \sqrt{3} H \tan( \alpha) \\ \\ \implies \sf \: 2 = \sqrt{3} \tan( \alpha ) \\ \\ \implies \sf \: \alpha = \tan {}^{ - 1} \bigg( \dfrac{2}{ \sqrt{3} } \bigg)$