- #1

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I was playing around with some equations regarding air resistance. I tried to calculate the height that is reached by an object that is projected vertically into the air. However something seems to go wrong when integrating.

Starting with the equation of motion

\begin{align*}

m\frac{dv}{dt}=-mg-kv^2.

\end{align*}

Setting \begin{align*}a=\sqrt{\frac{km}{g}},\\

v(0)=v_0.

\end{align*}

Then the solution to this differential equation is

\begin{align*}

v(t)=\frac{\tan{(\arctan{(av_0)}-gt})}{a}.

\end{align*}

Then the time it take to slow the object to a standstill(where it reaches maximum height) is

\begin{align*}

t_{end}=\frac{\arctan{(av_0)}}{g}.

\end{align*}

So the distance traveled in this time can be found by integrating the velocity function over this time.

\begin{align*}

h&=\int_0^{t_{end}}\frac{\tan{(\arctan{(av_0)}-gt})}{a}dt\\

&=\frac{1}{ga}(\ln{\cos{(arctan{(av_0)}-gt_{end})}}-\ln{\cos{(\arctan{(av_0)}}}.

\end{align*}

I did not even bother going through with it, it's going to come out negative.

I'm not actually sure this is the right antiderivative. Or maybe I lost a minus sign somewhere. I can't spot it.

Thanks :)