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Question: A projectile is fired with a speed v at an angle θ above the horizontal field. The coefficient of re...

A projectile is fired with a speed v at an angle θ above the horizontal field. The coefficient of restitution between the projectile and field is e. Find the position from the starting point when the projectile will land at its second collision.

A

e2u2sin⁡2θg\frac{e^{2}u^{2}\sin 2\theta}{g}

B

(1−e2)u2sin⁡2θg\frac{\left( 1 - e^{2} \right)u^{2}\sin 2\theta}{g}

C

(1−e2)u2sin⁡θcos⁡θg\frac{\left( 1 - e^{2} \right)u^{2}\sin\theta\cos\theta}{g}

D

(1+e)u2sin⁡2θg\frac{(1 + e)u^{2}\sin 2\theta}{g}

Answer

(1+e)u2sin⁡2θg\frac{(1 + e)u^{2}\sin 2\theta}{g}

Explanation

Solution

Vertical velocity after first collision = eu sin θ.

New Time of flight T' = 2eusin⁡θg\frac{2eu\sin\theta}{g},

R' = uxT'

x = R + R' = (1+e)u2sin⁡2θg\frac{(1 + e)u^{2}\sin 2\theta}{g}