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Q.

The range of the projectile projected at an angle of 15° with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45° with horizontal, then the range will be

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a

$50 \sqrt{2} m$

b

100 m

c

50 m

d

$100 \sqrt{2} m$

answer is C.

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Detailed Solution

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$\begin{array}{l} R = \frac{u^{2} \sin 2 θ}{g} \Rightarrow 50 = \frac{u^{2} \sin (2 \times 15)}{g} \\ R^{1} = \frac{u^{2} \sin (2 \times 45^{\circ})}{g} \\ \Rightarrow \frac{R^{1}}{50} = \frac{\sin 90^{\circ}}{\sin 30^{\circ}} = \frac{1}{1 / 2} = 2 \\ \Rightarrow R^{1} = 2 \times 50 = 100 m \end{array}$

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