Question

A ball is thrown from the ground with velocity u at an angle theeta with horizontal.the horizontal range of ball on ground is R .if the coefficient of restitution is e , then the horizontal range after the collision is.

Answer 1

Hi,

Answer:

The horizontal range after the collision is (e u² sin 2θ) /(g).

Explanation:

Given data:  

A ball is thrown with an initial horizontal velocity “u” with an angle “θ” with the horizontal ground. The horizontal range of the ball is given as “R” and the coefficient of restitution as “e”. Shown in fig.1 below.

To find: the horizontal range “R” after the collision of the ball with the ground.

After the collision of the ball with the ground, the horizontal component of the velocity along the ground remains the same as “ucosθ” and the vertical component velocity reverses and becomes as “usinθ”. Now, the ball will continue to make a projectile motion, as shown in fig.2 below, so, the vertical component of velocity will become “e” times.

Since, e = |(v₂-v₁) / (u₁-u₂)| = |v₁ / u₁| ……..[ ∵ v₂=u₂=0]

Or, v₁ = e u₁

∴the vertical component, v = e usinθ

The time taken to fligh at the highest point during the motion be “t”.

∴ v = u + at

Or, v = u – gt ……..[ ∵ acceleration is acting downwards]

also, final velocity at the highest point will be zero.

∴ 0 = u – gt

Or, u = gt

Or, e usinθ = gt ……[ here we are taking the vertical component of the velocity]

Or, t = (e u sin θ) / (g )

∴ Time of flight, T = 2 t = ( 2 e u sin θ) / (g) ……(i)

Therefore, the horizontal range after the collision will be

R = ucosθ * T

or, R = ucosθ * (2 e u sin θ) / (g)

or, R = (2 e ucos θ usin θ) / (g)

or, R = ( e u² sin 2θ ) / (g) …. [∵ 2 cosθsinθ = sin2θ]

Hope it helps!!!!!