Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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A Person Standing Near the Edge of the Top of a Building Throws Two Balls A And B. the Ball A Is Thrown Vertically Upward And B Is Thrown Vertically Downward with the Same Speed. - Physics

MCQ

A person standing near the edge of the top of a building throws two balls A and B. the ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed vA and the ball B this the ground with a speed vB. We have

Options

  • vA > vB

  • vA < vB

  • vA = vB

  • the relation between vA and vB depends on height of the building above the ground.

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Solution

vA = vB
Total energy of any particle = \[\frac{1}{2}m v^2 + mgh\]

Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.
At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal.

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 3 Rest and Motion: Kinematics
MCQ | Q 7 | Page 49
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