An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same object is th

Question

An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same object is then raised again to the same height h but this time is thrown downward with velocity v1 . It now reaches the ground with a new velocity v2 . How is v2 related to v1 ?

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Sapo 4 years 2021-07-28T23:21:28+00:00 1 Answers 335 views 0

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  1. kiet
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    2021-07-28T23:22:48+00:00
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    Answer:

    v_2=\sqrt{2}v_1

    Explanation:

    The velocity v₁ can be calculated with the kinematic formula:

    v_1^{2} =v_0^{2} +2gh

    Since the object is initially at rest, v₁ becomes:

    v_1=\sqrt{2gh}

    Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

    v_2^{2}=v_1^{2} +2gh

    Substituting v₁ in this expression and solving for v₂, we get:

    v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}

    Now, dividing v₂ over v₁, we get the expression:

    \frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1

    It means that v₂ is √2 times v₁.

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