Question

a projectile is fired at 30°and their maximum height is 4m another projectile with same speed is fired at 60°with maximum height is 16 m the range of 2nd projectile is​

Answer 1

        $\text{We know that max height of a projectile}(H_{max}) = \dfrac{u^2sin^2\theta}{2g}$

        $\text{Case I: at 30}^{\circ}$

$\implies \: 4 = \dfrac{u^2sin^230}{2g}$

$\implies \: 4 = \dfrac{u^2}{8g} \text{ ------ (i)}$

        $\text{Case II: at 60}^{\circ}$

$\implies \: 16 = \dfrac{u^2sin^260}{2g}$

$\implies \: 16 = \dfrac{3u^2}{8g} \text{ ------ (i)}$

        $\text{From (ii) we get }u = 8\sqrt{\dfrac{2g}{3}}$

        $\text{We know that Range}(R) = \dfrac{u^2sin2\theta}{g}$

$\implies \: R_2 = \dfrac{u^2sin120}g$

        $\text{We derived that }u = 8\sqrt{\dfrac{2g}{3}}$

$\implies \: R_2 = \dfrac{128g\times sin120}{3g}$

$\implies \: R_2 = \dfrac{128}{3} \times \dfrac{\sqrt3}{2}$

$\implies \: \boxed{R_2 = \dfrac{64}{\sqrt3} \: m}$