Summarized Problem: A baseball is thrown vertically up with speed $$v_o$$ and is subject to a quadratic drag with magnitude $$f(v) = cv^2$$. Write down the equation of motion for the upward journey (measuring y vertically up) and show that it can be rewritten as $$\dot{v} = —g[1 + (v/v_{\textrm{ter}})^2]$$. Show that the baseball's maximum height is $y_{\textrm{max}} = \frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{v^2_{\textrm{ter}} + v^2_o}{v^2_{\textrm{ter}}}\right)$

Solution: Writing the equation of motion is pretty straight forward: the two forces acting on the ball are the quadratic drag force and gravity. Therefore, $F = -f(v) - mg = -cv^2 - mg,$ meaning that $\dot{v} = -\frac{cv^2}{m} - g.$ We can then factor this equation: $\dot{v} = -g\left[1+ \frac{cv^2}{mg}\right].$ Now, to answer the first part of the question, we need to show that $$\frac{mg}{c} = v^2_{\textrm{ter}}$$. If the forces acting on an object in free fall are $F = cv^2 - mg,$ then the terminal velocity can only be reached when $$cv^2 = mg$$, meaning that $v^2_{\textrm{ter}} = \frac{mg}{c}.$ The next part of the question is asking us to find the maximum height of the ball in terms of its initial velocity using the handy relation $$\dot{v} = v\frac{dv}{dy}$$. Since we already have an equation for $$\dot{v}$$, this differential equation is pretty easy to solve. Once the variables are separated, we must find the definite integral from $$v_o$$ to $$v$$. The reason we must find the definite integral instead of the indefinite integral is because this equation is a little weird in the sense that instead of time being the variable, velocity is. Since the velocity is decreasing predictably (it's decreasing quadratically), finding the definite integral essentially gives us a proxy for how much time has passed, and subsequently the distance travelled.

For example, the only time (along the upward trajectory) at which $$v=v_o$$ is $$t=0$$, so if we evaluate the definite integral from $$v_o$$ to $$v_o$$, we'll obviously find that the baseball hasn't travelled any distance at all. Also, keep in mind that this equation only describes the upward trajectory of the ball. $\dot{v} = v\frac{dv}{dy}$ $-g\left[1+ \frac{v^2}{v^2_{\textrm{ter}}}\right] = v\frac{dv}{dy}$ $\frac{-g\left[1+ \frac{v^2}{v^2_{\textrm{ter}}}\right]}{vdv} = \frac{1}{dy}$ $\frac{vdv}{-g\left[1+ \frac{v^2}{v^2_{\textrm{ter}}}\right]} = dy$ $y = -\frac{1}{g} \int_{v_o}^{v} \frac{v}{\frac{v^2}{v^2_{\textrm{ter}}} + 1}dv$ Let $$u = \frac{v^2}{v^2_{\textrm{ter}}} + 1$$ and $$du = \frac{2}{v^2_{\textrm{ter}}}vdv$$ $y = -\frac{v^2_{\textrm{ter}}}{2g} \int_{v_o}^{v} \frac{1}{u}du$ $y = \left. -\frac{v^2_{\textrm{ter}}}{2g} \ln u \right|_{v_o}^{v}$ $y = \left. -\frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{v^2}{v^2_{\textrm{ter}}} + 1\right) \right|_{v_o}^{v}$ $y = -\frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{v^2}{v^2_{\textrm{ter}}} + 1\right) + \frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{v_o^2}{v^2_{\textrm{ter}}} + 1\right)$ Since we're looking for $$y_\textrm{max}$$, we're looking for the height when the velocity is 0, meaning that the ball is at the peak of its trajectory and about to head back towards the ground. Thus, we can evaluate the above equation at $$v=0$$ to find $$y_\textrm{max}$$. $y_{\textrm{max}} = -\frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{0^2}{v^2_{\textrm{ter}}} + 1\right) + \frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{v_o^2}{v^2_{\textrm{ter}}} + 1\right)$ $y_{\textrm{max}} = \frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{v_o^2}{v^2_{\textrm{ter}}} + 1\right)$ $y_{\textrm{max}} = \frac{v^2_{\textrm{ter}}}{2g} \ln \left(\frac{v_o^2 + v^2_{\textrm{ter}}}{v^2_{\textrm{ter}}}\right)$ And thus, we have shown that the peak of the trajectory that the question posed is correct.