Physics · Motion in Plane

JEE Advanced 2019 — Paper 2 — Question 10

A ball is thrown from ground at an angle θ\theta with horizontal and with an initial speed u0\mathrm{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 V1\mathrm{V}_{1}. After hitting the ground, the ball rebounds at the same angle θ\theta but with a reduced speed of u0/αu_{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 0.8 V10.8 \mathrm{~V}_{1}, the value of α\alpha is ____\_\_\_\_ —.

Question figure

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

Let 2u0sin⁡θ g=T0\frac{2 \mathrm{u}_{0} \sin \theta}{\mathrm{~g}}=\mathrm{T}_{0} and u0cos⁡θ=v1\mathrm{u}_{0} \cos \theta=\mathrm{v}_{1} (given)

Average velocity =u0cos⁡θT0+u0αcos⁡θ Tα+u0α2cos⁡θ T0α2T0+T0α+T0α2+……=\frac{\mathrm{u}_{0} \cos \theta T_{0}+\frac{\mathrm{u}_{0}}{\alpha} \cos \theta \frac{\mathrm{~T}}{\alpha}+\frac{\mathrm{u}_{0}}{\alpha^{2}} \cos \theta \frac{\mathrm{~T}_{0}}{\alpha^{2}}}{T_{0}+\frac{T_{0}}{\alpha}+\frac{T_{0}}{\alpha^{2}}+\ldots \ldots}

=u0cos⁡011+1α2+1α4+……….T0−1+1α+1α2+…..=v1αα+1=0.8 V=\frac{u_{0} \cos _{0} \frac{1}{1+\frac{1}{\alpha^{2}}+\frac{1}{\alpha^{4}}+\ldots \ldots \ldots .}}{\mathrm{T}_{0}-1+\frac{1}{\alpha}+\frac{1}{\alpha^{2}}+\ldots . .}=v_{1} \frac{\alpha}{\alpha+1}=0.8 \mathrm{~V}

α=4\alpha=4

Answer key and solution verified before publishing.

Practise Motion in Plane

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Advanced 2019
Paper
Paper 2
Subject
Physics
Chapter
Motion in Plane
Topic
Oblique and Horizontal Projectile Motion