JEE Advanced 2022 — Paper 1 — PHY-I solution

JEE Advanced2022Paper 1PHY-I

A projectile is fired from horizontal ground with speed $v$ and projection angle $θ$ . When the acceleration due to gravity is $g$ , the range of the projectile is $d$ . If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g^{'} = \frac{g}{0.81}$ , then the new range is $d^{'} = n d$ . The value of $n$ is _____.

Your answer

Solution

Original range: $d = \frac{v ^{2} sin 2 θ}{g}$ . Maximum height: $H = \frac{v ^{2} sin ^{2} θ}{2 g}$ .

At the apex, vertical speed is zero and horizontal speed is $v cos θ$ . From here the projectile falls through height $H$ under $g^{'} = g /0.81$ .

Time to reach ground from apex: $t = \frac{2 H}{g ^{'}} = \frac{2 H \cdot 0.81}{g} = 0.9 \frac{2 H}{g} = 0.9 \cdot \frac{v sin θ}{g}$ .

Horizontal distance in new region: $x = v cos θ \cdot t = 0.9 \cdot \frac{v ^{2} sin θ cos θ}{g} = 0.9 \cdot \frac{d}{2} = 0.45 d$ .

Total new range: $d^{'} = \frac{d}{2} + 0.45 d = 0.95 d$ . $∴ n = 0.95$ .

← Back to full paper

Practice more PHY-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →