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 $2 u$ , its range will be

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Solution

Since horizontal range,
$R = \frac{u^{2} sin 2 θ}{g}$

$∴ R \propto u^{2} sin 2 θ$

$\frac{R_{2}}{R_{1}} = {(\frac{u_{2}}{u_{1}})}^{2} (\frac{sin 2 θ_{2}}{sin 2 θ_{1}})$

Given:
$u_{1} = u, θ_{1} = 15^{\circ}, u_{2} = 2 u, θ_{2} = 45^{\circ}, R_{1} = R$
$\Rightarrow R_{2} = R_{1} {(\frac{2 u}{u})}^{2} (\frac{sin 90^{\circ}}{sin 30^{\circ}}) = 8 R$

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

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