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The horizontal range of a projectile fired at an angle of $15^{\circ}$ is 50 m. If it is fired with the same speed at an angle of $45^{\circ}$, its range will be
Explanation
Consider, projectile is fired at an angle $\theta$. According to question, $\theta = 15^{\circ}$ and $R = 50 m$ Range, $R = u^{2}\sin 2\theta/g$ $R = 50 m = u^{2}\sin(2 \times 15^{\circ})/g$ $50 \times g = u^{2}\sin 30^{\circ} = u^{2} \times (1/2) \Rightarrow 50 \times g \times 2 = u^{2}$ $u^{2} = 50 \times 9.8 \times 2 = 100 \times 9.8 = 980$ $u = \sqrt{980} = 31.304 m/s = 14\sqrt{5} ( \because g = 9.8 m/s^{2})$ Now, $\theta = 45^{\circ}$; $R = u^{2}\sin(2 \times 45^{\circ})/g = u^{2}/g = 100 m$