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Physics · Motion in Plane

JEE Main 2024 — 8 April, Shift 2 — Question 56

A body of mass M thrown horizontally with velocity vv from the top of the tower of height H touches the ground at a distance of 100 m from the foot of the tower. A body of mass 2 M thrown at a velocity v2\frac{v}{2} from the top of the tower of height 4 H will touch the ground at a distance of \qquad .m.

Answer: 100

Numerical answer — enter this value.

Step-by-step solution

100=v2H g;x=v22(4H)g=v2H g⇒x=100\begin{aligned} & 100=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{~g}}} ; \quad \mathrm{x}=\frac{v}{2} \sqrt{\frac{2(4 \mathrm{H})}{\mathrm{g}}}=v \sqrt{\frac{2 \mathrm{H}}{\mathrm{~g}}} & \Rightarrow \mathrm{x}=100 \end{aligned}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 2
Subject
Physics
Chapter
Motion in Plane
Topic
Oblique and Horizontal Projectile Motion

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