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Question

# A ball is projected with an initial speed u at an angle θ from the horizontal ground. The coefficient of restitution between the ball and the ground is e. Find the position from the starting point when the ball will land on the ground for the 2nd time.

A
e2u2sin2θg
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B
(1e2)u2sin2θg
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C
(1e)u2sinθcosθg
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D
(1+e)u2sin2θg
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Solution

## The correct option is D (1+e)u2sin2θgMotion of a particle projected at an angle θ with horizontal will be parabolic, and it is a symmetrical motion about y−axis. Range of the particle when it strikes the ground for 1st time: R1=u2sin2θg Just after collison, the y−component of velocity vy will change its magnitude & direction due to impulsive force from the ground in +ve y− direction. e=speed of separationspeed of approach ⇒e=vyusinθ ∴vy=eusinθ Now, this will be a second projectile motion as shown in figure below. New time of flight T′=2u⊥g=2eusinθg, Hence horizontal distance(Range) travelled by ball when it lands on the ground for 2nd time: R2=uxT′ ⇒R2=ucosθ×2eusinθg ∴R2=eu2sin2θg Hence, total distance travelled by the ball along the ground, when it lands for 2nd time: X=R1+R2 ∴X=u2sin2θg(1+e)

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