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    Topics || Problems

    If an object is shot vertically from the surface of the earth with an initial velocity of \(v\) feet per second, and if air resistance and other disturbing factors are neglected, it is found that \(s = vt - 0.5gt^2\), where \(s\) feet is the height of the object above the surface at the end of \(t\) seconds and \(g = 32\), approximately. (a) Solve for \(t\) in terms of \(s\) (b) If \(v=300 \text{fps}\), use (a) to find where \(s=450\) and \(s=0\).
    Solution:
    (a)

    \(s = vt -0.5gt^2\)

    \(2s =2vt -gt^2\)

    \( gt^2 -2vt = -2s \)

    \(t^2 - \frac{2v}{g}t = \frac{-2s}{g} \)

    \(t^2 -\frac{2v}{g}t + (\frac{v}{g})^2 = \frac{-2s}{g} + (\frac{v}{g})^2\)

    \( (t-\frac{v}{g})^2 = \frac{v^2 - 2sg}{g^2}\)

    \(t - \frac{v}{g} = \frac{\sqrt{v^2-2sg}}{g}\)

    \(t = \frac{v\pm\sqrt{v^2-2sg}}{g}\)

    (b)

    When \(s = 450\)

    \(t = \frac{300 + \sqrt{300^2-2(450)(32)}}{32}\)

    \(t = 17.11 \text{sec}\)

    \(t = \frac{300 - \sqrt{300^2-2(450)(32)}}{32}\)

    \(t = 1.64 \text{sec}\)

    When \(s = 0\)

    \(t = \frac{300 + \sqrt{300^2-2(0)(32)}}{32}\)

    \(t = 18.75 \text{sec}\)

    \(t = \frac{300 - \sqrt{300^2-2(0)(32)}}{32}\)

    \(t = 0 \text{sec}\)