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A projectile is thrown with speed v₀ at an angle θ with the horizontal plane. It hits a vertical elastic wall after crossing the maximum height and rebounds with same horizontal velocity and finally falls at a point on the horizontal plane just below the highest point on the trajectory. The total horizontal distance travelled by the projectile is

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$\frac{v_{θ}^{2} \sin 2 θ}{2 g}$

$\frac{v_{0}^{3} \sin 2 θ}{4 g}$

$\frac{v_{θ}^{2} \sin θ}{g}$

$\frac{v_{0}^{2} \sin 2 θ}{g}$

answer is D.

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Detailed Solution

Since the collision is elastic, dotted line is the mirror image of the actual trajectory after hitting the wall. So, total horizontal distance = range = $\frac{v_{0}^{2} \sin 2 θ}{g}$

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