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19 June, 15:53

A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Neglect air friction)

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  1. 19 June, 17:35
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    4452.5 J.

    Explanation:

    The diver have both kinetic and potential energy.

    Ek = 1/2mv² ... Equation 1

    Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

    Given: m = 65 kg, v = 6.4 m/s.

    Substitute into equation 1

    Ek = 1/2 (65) (6.4²)

    Ek = 1331.2 J.

    Also,

    Ep = mgh ... Equation 2

    Where Ep = Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

    Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

    Substitute into equation 2.

    Ep = 65 (4.9) (9.8)

    Ep = 3121.3 J.

    Note: When she hits the water, the potential energy is converted to kinetic energy.

    E = Ek+Ep

    Where E = Kinetic energy of the diver when she hits the water.

    E = 1331.2+3121.3

    E = 4452.5 J.
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