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Thread: velocity of a falling ball help

  1. #1
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    velocity of a falling ball help

    In our simple model of the fight of a ball, the height
    y above the foor at time t is

    $\displaystyle y = V_{n}t-\frac{1}{2}gt^{2} $



    where g is the acceleration due to gravity and vn is the velocity immediately after the nth
    bounce.

    The maximum height of the ball is reached when dy/dt = 0
    Find the time
    tmax at which this maximum occurs and the maximum height
    ymax
    that the ball reaches.


    I am really not sure how to do this, since I do not know what vn is?

    $\displaystyle \frac{dy}{dt} = V_{n} -gt = 0 $

    $\displaystyle V_{n} = gt $

    dont know how to work ourt $\displaystyle y_{max} $

    and $\displaystyle t_{max} $



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  2. #2
    MHF Contributor
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    Re: velocity of a falling ball help

    $\displaystyle t_{max}$ is the time when the derivative is zero, and you found the equation $\displaystyle V_n=gt$ for that. So you solve for t and get $\displaystyle t_{max}=\frac{V_n}{g}$. Then you plug in to $\displaystyle y = V_{n}t-\frac{1}{2}gt^{2}$ to get the height at time $\displaystyle t_{max}$, which is $\displaystyle y_{max}$.

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