JEE Advanced 2019 · Paper 2 · Physics Q9

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PhysicsNumerical+3Question 9 of 54

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, the 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 ______.

A ball bounces along level ground in successive parabolic arcs; each bounce leaves at the same angle θ, with speeds u₀, u₀/α, u₀/α², …, u₀/αᵐ.

Your answer

Answer and worked solution

Official answer: 4

First flight. The average velocity of a projectile landing at its launch height is its horizontal velocity:
$V_1=\frac RT=u_0\cos\theta.$

All flights. Each bounce divides the speed by $\alpha$ , so ranges ( $\propto u^2$ ) shrink by $\alpha^2$ and times ( $\propto u$ ) by $\alpha$ :
$\sum R=R\frac{\alpha^2}{\alpha^2-1},\qquad\sum T=T\frac\alpha{\alpha-1}.$

Overall average velocity.
$\bar V=V_1\frac{\alpha^2/(\alpha^2-1)}{\alpha/(\alpha-1)}=V_1\frac\alpha{\alpha+1}=0.8V_1\implies\alpha=4.$
Answer: 4

Topics:Projectile motion Motion in a plane

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JEE Advanced 2019 · Paper 2, Physics question 9