Physics · Motion in Plane

JEE Main 2025 — 29 January, Morning Shift — Question 44

The maximum speed of a boat in still water is 27 km/h27 \mathrm{~km} / \mathrm{h}.

Now this boat is moving downstream in a river flowing at 9 km/h9 \mathrm{~km} / \mathrm{h}.

A man in the boat throws a ball vertically upwards with speed of 10 m/s10 \mathrm{~m} / \mathrm{s}.

Range of the ball as observed by an observer at rest on the river bank, is \qquad cm . (Take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} )

Answer: 2000

Numerical answer — enter this value.

Step-by-step solution

v→b=9+27=36 km/hr\overrightarrow{\mathrm{v}}_{\mathrm{b}}=9+27=36 \mathrm{~km} / \mathrm{hr} →⟶\xrightarrow{\longrightarrow}

v→b=36×100036000=10 m/sec\overrightarrow{\mathrm{v}}_{\mathrm{b}}=36 \times \frac{1000}{36000}=10 \mathrm{~m} / \mathrm{sec}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Motion in Plane
Topic
Oblique and Horizontal Projectile Motion
The maximum speed of a boat in still water is 27 km / h . Now this… | JEE Main 2025 PYQ with Solution · DhiX AI