A body is thrown vertically upward with initial velocity

A body is thrown vertically upward with initial velocity $v_0$. Taking air resistance into account, its velocity is $v=c\tan\left(-\dfrac{gt}{c}+\arctan\dfrac{v_0}{c}\right)$, where $c$ is a constant. Find the maximum height reached.

Step 1: Time to reach maximum height

Setting $v=0$ requires the tangent's argument to vanish (its first zero, since the body is still ascending):

\[-\frac{gt_{\max}}{c}+\arctan\frac{v_0}{c}=0 \quad\Longrightarrow\quad \boxed{t_{\max}=\frac{c}{g}\arctan\frac{v_0}{c}}.\]

Step 2: Set up the height integral

\[h_{\max}=\int_0^{t_{\max}}v\,dt=\int_0^{t_{\max}}c\tan\left(-\frac{gt}{c}+\arctan\frac{v_0}{c}\right)dt.\]

Step 3: Substitute $u=-\dfrac{gt}{c}+\arctan\dfrac{v_0}{c}$

Then $du=-\dfrac{g}{c}dt$; at $t=0$, $u=\arctan(v_0/c)$, and at $t=t_{\max}$, $u=0$. So

\[h_{\max}=\frac{c^2}{g}\int_0^{\arctan(v_0/c)}\tan u\,du=\frac{c^2}{g}\Big[\ln|\sec u|\Big]_0^{\arctan(v_0/c)}=\frac{c^2}{g}\ln\sec\!\left(\arctan\frac{v_0}{c}\right).\]

Step 4: Simplify $\sec(\arctan(v_0/c))$

With $\theta=\arctan(v_0/c)$, $\tan\theta=v_0/c$, and $\sec^2\theta=1+\tan^2\theta=1+v_0^2/c^2$, so $\sec\theta=\sqrt{1+v_0^2/c^2}$.

Step 5: Final result

\[h_{\max}=\frac{c^2}{g}\ln\sqrt{1+\frac{v_0^2}{c^2}}=\frac{c^2}{2g}\ln\!\left(1+\frac{v_0^2}{c^2}\right).\]

\[\boxed{h_{\max}=\frac{c^2}{2g}\ln\!\left(1+\frac{v_0^2}{c^2}\right)}\]

Step 6: Check against the drag-free limit

As $c\to\infty$ (air resistance vanishes), $\ln(1+v_0^2/c^2)\approx v_0^2/c^2$, so $h_{\max}\to\dfrac{v_0^2}{2g}$, recovering the classical result of Problem 162 as it must.