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28 December, 20:31

You and your friends find a rope that hangs down 15 m from a high tree branch right at the edge of a river. You find that you can run, grab the rope, swing out over the river, and drop into the water. You run at 2.0 m/s and grab the rope, launching yourself out over the river.

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  1. 28 December, 22:11
    0
    Vertical angle=9.5°

    Time to reach maximum height = 1.24seconds

    Step-by-step explanation:

    The maximum distance x is reached when rope swings to a maximum height, maximum height is gotten from the equation KE = 1/2MU^2 = MgH = MgL = Mgl (1 - cos theta.

    Where theta is the vertical angle, U=2m/s and L = 15m

    2gl - 2glcostheta=U^2

    2gl U^2=2gl cos theta

    But theta = Acos (1 - U^2) / 2gl

    Acos (1- / (2*9.8*15)

    Theta = 9.5°

    Maximum height li given by

    Lsintheta

    But X = UxT

    T=X/UX = Lsintheta/U = 15sin9.5 / 2 = 1.24seconds
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