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26 July, 01:47

If a rock climber accidentally drops a 56.0 g piton from a height of 295 m, what would its speed be just before striking the ground? Ignore the effects of air resistance.

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  1. 26 July, 02:07
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    The speed of the piton before hitting the ground = 76.04 m/s

    Explanation:

    From the law of conservation of energy,

    The kinetic energy of the piton = potential energy of the piton

    Note:Before striking the ground, the potential energy of the piton is converted to kinetic energy, with no energy lost to air resistance.

    1/2mv² = mgh ... Equation 1

    Where m = mass of the piton, v = speed of the piton, h = height, g = acceleration due to gravity.

    making v the subject of the equation above,

    v = √ (2gh) ... Equation 2

    Given: h = 295 m, g = 9.8 m/s²

    Substituting into equation 2

    v = √ (2*9.8*295)

    v = √5782

    v = 76.04 m/s

    Thus the speed of the piton before hitting the ground = 76.04 m/s
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