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Physics

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It takes 0.220 s for a dropped object to pass a window that is 1.14 m tall. From what height above the top of the window was the object released?

  • Physics - ,

    Let d be the distance from the drop point to the top of the window.
    The time to reach the top is
    t = sqrt(2d/g)
    The time to reach the bottom is
    t' = t + 0.220 = sqrt[2(d+1.14)/g]
    = sqrt(2d/g) + 0.22

    It's rather messy, but there you have one equation in the single unknown, d. Solve for d.

  • Physics - ,

    Let:
    d1 = distance from drop to top of window
    d2 = distance from drop to bottom of window
    t1 = time from ball drop to top of window
    t2 = time from ball drop to bottom of window
    vi = initial velocity
    g = acceleration due to gravity
    Knowing the equation (d = vi*t + (a*t^2)/2)
    d1 = (g*t1^2)/2
    d2 = (g(t1+.22)^2)/2
    d2 = d1 + 1.14
    SO:
    (g(t1+.22)^2)/2 = (g*t1^2)/2 + 1.14
    Solve for t1.
    Use t1 = sqrt(2*d1/g)
    Solve for d1 and you will have the distance from the drop to the top of the window.

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