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14 January, 01:07

An archer puts a 0.4 kg arrow to the bowstring. An average force of 190.4 N is exerted to draw the string back 1.47 m. The acceleration of gravity is 9.8 m/s². Assuming no frictional loss, with what speed does the arrow leave the bow? Answer in units of m/s. If the arrow is shot straight up, how high does it rise? Answer in units of m.

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  1. 14 January, 03:18
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    v = 37.4 m/s, h = 71.39m

    Explanation:

    To find the velocity given:

    m = 0.4 kg

    F = 190.4 N

    d = 1.47 m

    g = 9.8 m/s^2

    So use the equation of work to solve the kinetic energy

    W = F * d = 190.4 N * 1.47m

    W = 279.88 J

    Ke = 1 / 2 * m * v^2

    v = √2*Ke / m = √ 2 * 279.88 / 0.4 kg

    v = 37.4 m/s

    Now to find the high to rise can use the conserved law so:

    Ke = Pe

    279.88 = m*g*h

    Solve to h'

    h = 279.88 / 0.4kg * 9.8m/s^2

    h = 71.39 m
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