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6.3C: Body thrown vertically upwards with initial speed v₀
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/06%3A_Motion_in_a_Resisting_Medium/6.03%3A_Uniformly_Accelerated_Motion/6.3C%3A_Body_thrown_vertically_upwards%2C_initial_speed_(v_0)
$v = -\hat{v} + (v_{0}+\hat{v})e^{- \gamma t} \tag{6.3.25}\label{eq:6.3.25}$ The second time integral (obtained by writing $v$ as $\frac{dy}{dt}$ in Equation $\ref{eq:6.3.25}$ ) and the spa... $v = -\hat{v} + (v_{0}+\hat{v})e^{- \gamma t} \tag{6.3.25}\label{eq:6.3.25}$ 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 $v = v_{0} - \gamma y -\hat{v}\ln \left(\frac{\hat{v}+v_{0}}{\hat{v}+v}\right). \tag{6.3.28}\label{eq:6.3.28}$

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