Physics · Capacitors and R-C Circuits

JEE Advanced 2021 — Paper 1 — Question 33

A projectile is thrown from a point $O$ on the ground at an angle $45^{\circ}$ from the vertical and with a speed $5 \sqrt{2} \mathrm{~m} / \mathrm{s}$ . The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O . The acceleration due to gravity $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ .

The value of $x$ is $\text{ }\!\!~\!\!\text{ }$ .

Answer: 7.50

Numerical answer — enter this value.

Step-by-step solution

Velocity of projectile at highest point $5\text{ }\!\!~\!\!\text{ m}/\text{s}\overset{\text{}}{\mathop{\text{i}}}\,$

Since, there is no external force in horizontal direction so by conservation of momentum

           $m\left( 5 \right)=\frac{m}{2}\left( 0 \right)+\frac{m}{2}\left( v \right)$

$\text{vec{v}}=10\text{ }\!\!~\!\!\text{ m}/\text{s}\overset{\text{}}{\mathop{\text{i}}}\,$ Distance covered by second mass before landing = $\frac{\text{ }\!\!~\!\!\text{ Range }\!\!~\!\!\text{ }}{2}+10\left( \text{t} \right)=7.5\text{ }\!\!~\!\!\text{ m}$ y

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Exam: JEE Advanced 2021
Paper: Paper 1
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Charging and Discharging of R-C Circuits

Answer: 7.50

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