The range of the projectile projected at an angle of 15° with the horizontal is 50 m. If the projectile is projected with the same velocity at an angle of 45° with the horizontal, then its range will be [2023]

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Solution

Q.
The range of the projectile projected at an angle of 15° with the horizontal is 50 m. If the projectile is projected with the same velocity at an angle of 45° with the horizontal, then its range will be [2023]

1 50 m

2 $50 \sqrt{2}$ m

3 100 m

4 $100 \sqrt{2}$ m

Ans.
(3)

$R = \frac{v^{2} \sin 2 θ}{g}$

$R \propto \sin (2 θ)$

$\frac{R_{1}}{R_{2}} = \frac{\sin (2 θ_{1})}{\sin (2 θ_{2})} = \frac{\sin (2 \times 15 °)}{\sin (2 \times 45 °)} = \frac{\sin 30 °}{\sin 90 °}$

$\frac{50}{R_{2}} = \frac{1}{2}$

$R_{2} = 100 m$

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