Throwing a Ball Up: Motion & Time to Max Height

[Dustinsfl]

Consider an object that is thrown vertically up with initial speed $v_0$ in a linear medium.
Measuring $y$ upward from the point of release, write expressions for the object's velocity $v_y(t)$ and position $y(t)$.
The equation of motion is
\begin{alignat*}{3}
m\dot{v}_y & = & -mg - bv_y\\
& = & -b\left(\frac{mg}{b} + v_y\right)\\
\dot{v}_y & = & -\frac{b}{m}\left(v_{\text{ter}} + v_y\right)\\
& = & -\frac{1}{\tau}\left(v_{\text{ter}} + v_y\right)\\
\int_{v_0}^{v_y}\frac{dv_y'}{v_{\text{ter}} + v_y'} & = & -\frac{1}{\tau}\int_0^t dt'\\
\ln\left(\frac{v_{\text{ter}} + v_y}{v_{\text{ter}} + v_0}\right) & = & -\frac{t}{\tau}\\
v_y(t) & = & (v_{\text{ter}} + v_0)e^{-t/\tau} - v_{\text{ter}}\\
& = & v_0e^{-t/\tau} - v_{\text{ter}}(1 - e^{-t/\tau})
\end{alignat*}
Next, we need to solve for $y(t)$ where $\dot{y}(t) = v_y(t)$.
\begin{alignat*}{3}
\dot{y}(t) & = & v_0e^{-t/\tau} - v_{\text{ter}}(1 - e^{-t/\tau})\\
\int_{y_0 = 0}^ydy' & = & \int_0^t(v_{\text{ter}} + v_0)e^{-t'/\tau}dt' - \int_0^tv_{\text{ter}} dt'\\
y(t) & = & -tv_{\text{ter}} + \tau(v_{\text{ter}} + v_0)(1 - e^{-t/\tau})\\
\end{alignat*}
The time it reaches $y_{\max}$ is $t = \tau\ln\left(1 + \frac{v_0}{v_{\text{ter}}}\right)$ correct?

 

[kuruman]

Correct.