Physics · Motion in Plane

JEE Advanced 2018 — Paper 2 — Question 7

A ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is $\_\_\_\_$ -.

Answer: 30

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{H}_{0}=\frac{\left(\mathrm{v}_{0} \sin 45^{\circ}\right)^{2}}{2 \mathrm{~g}}=\frac{\mathrm{v}_{0}^{2}}{4 \mathrm{~g}}$

$\frac{\mathrm{v}_{0}^{2}}{4 \mathrm{~g}}=120 …(i)$

Now, $\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \times \frac{1}{2} \mathrm{mv}_{0}^{2}$

$\mathrm{v}=\frac{\mathrm{v}_{0}}{\sqrt{2}}$

Now, $H=\frac{\left(v \sin 30^{\circ}\right)^{2}}{2 \mathrm{~g}}=\frac{\mathrm{v}^{2}}{8 \mathrm{~g}}=\frac{\mathrm{v}_{0}^{2}}{16 \mathrm{~g}}=\frac{\mathrm{H}_{0}}{4}=30 \mathrm{~m}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Physics
Chapter: Motion in Plane
Topic: Oblique and Horizontal Projectile Motion