Question

a projectile is fired in such way that its horizontal range is equal to three times its maximum height. What is the angle of projection?​

Answer 1

Explanation:

Range=3×max height

$\frac{u {}^{2} \times sin2 \alpha }{g} = 3 \times \frac{u {}^{2} \times \sin \alpha {}^{2} }{2g} \\ = \sin(2 \alpha ) = \frac{3sin \alpha {}^{2} }{2} \\ = sin \alpha cos \alpha = \frac{3sin \alpha { }^{2} }{2} \\ \frac{2}{3} = tan \alpha \\ \alpha = tan {}^{ - 1} ( \frac{2}{3} )$

hope it helps:)