Kinematics

JEE Main level, numerical answer, medium

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Question 141 of 308

The maximum speed of a boat in still water is \(27 \mathrm{~km} / \mathrm{h}\). Now this boat is moving downstream in a river flowing at \(9 \mathrm{~km} / \mathrm{h}\). A man in the boat throws a ball vertically upwards with speed of \(10 \mathrm{~m} / \mathrm{s}\). Range of the ball as observed by an observer at rest on the river bank, is _______ cm. (Take \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\))

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Answer: 2000

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



v→b=36×100036000=10 m/sec Time of flight =2×1010=2sec Range =10×2=20 m=2000 cm\begin{aligned} & \overrightarrow{\mathrm{v}}_{\mathrm{b}}=36 \times \frac{1000}{36000}=10 \mathrm{~m} / \mathrm{sec} \\ & \text { Time of flight }=\frac{2 \times 10}{10}=2 \mathrm{sec} \\ & \text { Range }=10 \times 2=20 \mathrm{~m}=2000 \mathrm{~cm}\end{aligned}

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