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The range of the projectile projected at an angle of 15$$^\circ$$ with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45$$^\circ$$ with horizontal, then its range will be
The range of a projectile is given by: $$R = \frac{u^2 \sin 2\theta}{g}$$
At $$\theta = 15°$$: $$R_1 = \frac{u^2 \sin 30°}{g} = \frac{u^2}{2g} = 50$$ m
At $$\theta = 45°$$: $$R_2 = \frac{u^2 \sin 90°}{g} = \frac{u^2}{g}$$
From the first equation: $$\frac{u^2}{g} = 100$$ m
Therefore $$R_2 = 100$$ m.
The correct answer is Option 1: 100 m.
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