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6.3C: Body thrown vertically upwards, initial speed \(v_{0}\)

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    8889
  • If we measure \( y\) upwards from the ground, the equation of motion is

    \[ \ddot{y}= - g - \gamma v = -\gamma (\hat{v}+v). \tag{6.3.24}\label{eq:6.3.24} \]

    The first time integral is

    \[ v = -\hat{v} + (v_{0}+\hat{v})e^{- \gamma t} \tag{6.3.25}\label{eq:6.3.25} \]

    and this is shown in Figure VI.9.

    It reaches a maximum height after time \( T\), when \( v=0\)(at which time the acceleration is just \( -g\)):

    \[ t + \frac{1}{\gamma}\ln(1 + \frac{v_{0}}{\hat{v}}). \tag{6.3.26}\label{eq:6.3.26} \]

    The second time integral (obtained by writing \( v\) as \( \frac{dy}{dt}\) in Equation \(\ref{eq:6.3.25}\)) and the space integral (obtained by writing \( \ddot{y} \) as \( v\frac{dv}{dy}\) in the equation of motion) require some patience, but the results are

    \[ y = \frac{(v_{0}+\hat{v}}{\gamma}(1-e^{-\gamma t}- \hat{v}t, \tag{6.3.27}\label{eq:6.3.27} \]

    \[ v = v_{0} - \gamma y -\hat{v}\ln(\frac{\hat{v}+v_{0}}{\hat{v}+v}). \tag{6.3.28}\label{eq:6.3.28} \]

    These are illustrated in Figures VI.10 and VI.11.

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