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

7 R_{1}

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8 R_{1}

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9 R_{1}

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10 R_{1}

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Solution

The correct option is C $8 R_{1}$
$R = \frac{u^{2} sin 2 θ}{g} ∴ R \infty u^{2} sin 2 θ$
$\frac{R_{2}}{R_{1}} = {(\frac{u_{2}}{u_{1}})}^{2} (\frac{s i n 2 θ_{2}}{sin 2 θ_{1}})$
$\Rightarrow R_{2} = R_{1} {(\frac{2 u}{u})}^{2} (\frac{sin 90^{o}}{sin 30^{o}}) = 8 R_{1}$

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