Question

A ball is thrown from ground at an angle $\theta$ with horizontal and with an initial speed $u_0$. For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is $V_1$. After hitting the ground, ball rebounds at the same angle $\theta$ but with a reduced speed of $u_0/\alpha$. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is $0.8V_1$, the value of $\alpha$ is (JEE Adv. 2019)

Answer 1

4

Answer 2

Explanation:

1. For a projectile thrown with speed u₀ and angle θ, the range is R = (u₀² sin2θ)/g and flight time T = (2u₀ sinθ)/g. Their ratio gives the average velocity of the first throw:
     V₁ = R/T = u₀ cosθ.

2. For subsequent bounces, the speeds reduce by u₀/α, u₀/α², …, so:   - Range for the n-th bounce: Rₙ = (u₀² sin2θ)/(g α^(2(n-1))).   - Time for the n-th bounce: Tₙ = (2u₀ sinθ)/(g α^(n-1)).

3. Total horizontal displacement = (u₀² sin2θ)/g × [1 + 1/α² + 1/α⁴ + …]
     = (u₀² sin2θ)/g × 1/(1 – 1/α²).

4. Total time = (2u₀ sinθ)/g × [1 + 1/α + 1/α² + …]
     = (2u₀ sinθ)/g × 1/(1 – 1/α).

5. Therefore, the overall average velocity is
     V_avg = (Total displacement)/(Total time)
       = (u₀² sin2θ/g × 1/(1 – 1/α²)) / (2u₀ sinθ/g × 1/(1 – 1/α))
       = u₀ cosθ / (1 + 1/α)
       = V₁/(1 + 1/α).

6. Given that V_avg = 0.8V₁, we have:
     V₁/(1 + 1/α) = 0.8V₁.
     Cancelling V₁, we get
     1/(1 + 1/α) = 0.8
     => 1 + 1/α = 1/0.8 = 1.25
     => 1/α = 0.25
     => α = 4.