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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

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प्रश्न

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

पर्याय

  • vA > vB

  • vA < vB

  • vA = vB

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

MCQ
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उत्तर

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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पाठ 3: Rest and Motion: Kinematics - MCQ [पृष्ठ ४९]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 3 Rest and Motion: Kinematics
MCQ | Q 7 | पृष्ठ ४९
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