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14 September, 20:16

A gymnast practices two dismounts from the high bar on the uneven parallel bars. during one dismount, she swings up off the bar with an initial upward velocity of + 4.0 m/s. in the second, she releases from the same height but with an initial downward velocity of - 3.0 m/s. what is her acceleration in each case? how do the final velocities of the gymnast as she reaches the ground differ?

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  1. 14 September, 21:03
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    Note:

    The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).

    It is not provided in the question, so the standard height is assumed.

    g = 9.8 m/s², acceleration due to gravity.

    Note that the velocity and distance are measured as positive upward.

    Therefore the floor is at a height of h = - 2.8 m.

    First dismount:

    u = 4.0 m/s, initial upward velocity.

    Let v = the velocity when the gymnast hits the floor.

    Then

    v² = u² - 2gh

    v² = 16 - 2*9.8 * (-2.8) = 70.88

    v = 8.42 m/s

    Second dismount:

    u = - 3.0 m/s

    v² = (-3.0) ² - 2*9.8 * (-2.8) = 63.88 m/s

    v = 7.99 m/s

    The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.

    Answer:

    First dismount:

    Acceleration = 9.8 m/s² downward

    Landing velocity = 8.42 m/s downward

    Second dismount:

    Acceleration = 9.8 m/s² downward

    Landing velocity = 7.99 m/s downward

    The landing velocities differ by 0.43 m/s.
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