A projectile thrown with an initial speed $u$ and angle of projection $15^{\circ}$ to the horizontal has a range $R$. If the same projectile is thrown at an angle of $45^{\circ}$ to the horizontal with speed $2u$, its range will be

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Solution

Since horizontal range,
$R=\dfrac{u^2\sin2\theta}{g}$

$\therefore~R ~\propto~u^2\sin2\theta$

$ \dfrac{R_2}{R_1}=\left ( \dfrac{u_2}{u_1} \right )^2 \left ( \dfrac{\sin 2\theta_2}{\sin 2\theta_1} \right )$

Given:
$u_1=u, ~~~~\theta_1=15^{\circ},~u_2=2u,~~~~ \theta_2=45^{\circ},R_1=R$
$ \Rightarrow R_2 = R_1 \left ( \dfrac{2u}{u} \right )^2 \left ( \dfrac{\sin90^{\circ}}{\sin30^{\circ}} \right )=8R$

hence, option $(\text C)$ is the right choice.

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A projectile thrown with an initial speed \(u\) and angle of projection \(15^{\circ}\) to the horizontal has a range \(R\). If the same projectile is thrown at an angle of \(45^{\circ}\) to the horizontal with speed \(2u\), its range will be