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Physics · Motion in Plane

JEE Advanced 2022 — Paper 1 — Question 26

A projectile is fired from horizontal ground with speed v and projection angle θ\theta.

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′=g0.81g^{\prime}=\frac{g}{0.81},

then the new range is d′=nd\mathrm{d}^{\prime}=\mathrm{nd}. The value of nn is \qquad .

Answer: 0.95

Numerical answer — enter this value.

Step-by-step solution

d′=d2+2Hg′vcos⁡θ=d2+2 g′v2sin⁡2θ2 gvcos⁡θ\mathrm{d}^{\prime}=\frac{\mathrm{d}}{2}+\sqrt{\frac{2 \mathrm{H}}{\mathrm{g}^{\prime}}} \mathrm{v} \cos \theta=\frac{\mathrm{d}}{2}+\sqrt{\frac{2}{\mathrm{~g}^{\prime}} \frac{\mathrm{v}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}} \mathrm{v} \cos \theta d′=d2+V2sin⁡θcos⁡θ g g0.81=d2+d2(910)=19 d20\mathrm{d}^{\prime}=\frac{\mathrm{d}}{2}+\frac{\mathrm{V}^{2} \sin \theta \cos \theta}{\sqrt{\mathrm{~g} \frac{\mathrm{~g}}{0.81}}}=\frac{\mathrm{d}}{2}+\frac{\mathrm{d}}{2}\left(\frac{9}{10}\right)=\frac{19 \mathrm{~d}}{20}

d′=0.95 d\mathrm{d}^{\prime}=0.95 \mathrm{~d} n=0.95\mathrm{n}=0.95

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Physics
Chapter
Motion in Plane
Topic
Oblique and Horizontal Projectile Motion

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