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30 April, 10:30

A ball with mass 0.13 kg is thrown upward with initial velocity 20 m/s from the roof of a building 30 m high. neglect air resistance. (a computer algebra system is recommended. use g = 9.8 m/s2 for the acceleration due to gravity. round your answers to one decimal place.) (a) find the maximum height above the ground that the ball reaches.

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  1. 30 April, 12:26
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    We need to apply the following equation

    Vy ^2 = Voy^2 + 2 ay (y - yo)

    Where Vy is the final speed of the ball (which will be equal to zero at maximun height)

    Voy is the initial speed

    a is the acceleration (equal to - 9.8 m/s2, as it goes against the upward movement)

    yo is the initial reference position (yo = 30m from the ground)

    y = yo + [ (-Voy^2) / 2ay] = 30 m + - (20[m/s]) ^2 / - 2 (9.8[m/s^2])

    y = 30[m] + 20.408 [m] = 50.408 m = 50.4 m
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