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Thread: Free fall

  1. #1
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    Post Free fall

    Equation given -- s=6.1t^2, t is seconds.
    Object dropped from 75 feet. Find the speed of the object at t = 3.

    one question- is the 75 feet important? Or do you just plug in the 3 for t to get 54.9 ft/sec?

    Any help is appreciated!
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  2. #2
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    In the problem speed of the object is required.
    Speed v = ds/dt.
    In the problem 75 feet is necessary. Otherwise the object may reach the ground before 3 seconds.
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  3. #3
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    ok..then how would I set this problem up?
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  4. #4
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    They are a lot of things wrong with the original post. First, let's look at the vertical motion equations on Earth:
    $\displaystyle s(t) = -16t^2 + v_0 t + s_0$
    $\displaystyle v(t) = -32t + v_0$

    $\displaystyle t$ = time (I'll assume it's in seconds)
    $\displaystyle s_0$ = initial height
    $\displaystyle v_0$ = initial velocity
    $\displaystyle s(t)$ = height after t sec
    $\displaystyle v(t)$ = velocity after t sec

    The equation you've given us is
    $\displaystyle s = 6.1t^2$
    Does up => positive and down => negative? Or the other way around? If it is the former, then this equation is for an object that floats upwards. (In other words, is there a negative sign missing?)

    Since the object is being dropped, the initial velocity is zero, so it's okay that there is no t term in the equation you gave us. But there is mention of the initial height at all. So the question is, is this the equation you meant to give us?
    $\displaystyle s(t) = -6.1t^2 + 75$

    Or do you just plug in the 3 for t to get 54.9 ft/sec?
    Nope. The equation above looks like a position equation, not a velocity equation. Using the equation as originally given, plugging 3 in for t gives us a height of 54.9 feet, NOT a velocity of 54.9 ft/sec.


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