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The drawing shows a skateboarder moving at 7.59 m/s along a horizontal section of a track that is slanted upward by 55.0 ° above the horizontal at its end, which is 0.741 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.

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    Hmmm. The way I see it, kinetic energy is lost during her rise up the ramp. So, at the moment of takeoff, her KE is

    1/2 mv^2 - 1/2 mgh
    = m(1/2 * 7.59^2 - 1/2 * 9.8 * 0.741)
    = 25.17m

    Now, that means that her speed is then 7.10 m/s

    The vertical component of that is 5.81 m/s, so counting from the end of the track,

    H(t) = 5.81t - 4.9t^2

    which has a maximum at (.59,1.72)

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