a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

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a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

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Hải Đăng 3 years 2021-08-28T13:17:54+00:00 1 Answers 9 views 0

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  1. Thunguyet
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    2021-08-28T13:19:25+00:00
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    Answer:

    Approximately 7.0\; \rm m \cdot s^{-1}.

    Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

    Explanation:

    Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

    For an object that is accelerating constantly,

    v^2 - u^2 = 2\, a \, x,

    where

    • u is the initial velocity of the object,
    • v is the final velocity of the object.
    • a is its acceleration, and
    • x is its displacement.

    In this case, x is the same as the change in the ball’s height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

    v^2 - u^2 = 2\, g \cdot h.

    In this case, v would be the velocity of the ball just before it hits the ground. Solve for

    v^2 = 2\, a\, x + u^2.

    \begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

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