I have this problem:

"prove that the area under the trajectory described by a parabolic shot that has:

$f(x)=\tan(\theta)x - (\frac{g}{2v^2\cos^2(\theta)})x^2$



is defined as:


I have tried to change x in order to convert $f(x)$ to a function of $\theta$ but i end up everytime with $f(\theta)=0$

  • $\begingroup$ Please review how the trajectory has been arrived at.. what is asked is the difference with and without gravity term, when the shot goes in a straight line without gravity. $\endgroup$ – Narasimham Sep 3 '15 at 3:37

This is a straight-forward integral. $\theta$ is fixed for the projectile. If $\theta=0$ then the problem does not make much sense. If $\theta=\pi/2$ then the area is undefined. So it makes sense to assume $\theta\in(0,\pi/2)\cup(\pi/2,\pi)$

$$f(x)=0 \iff x=0 \ \ \vee \ \ x=\dfrac{2v^2\cos^2\theta\tan\theta}{g}$$

And so,

$$A(\theta) = \int_0^{2g^{-1}v^2\cos^2\theta\tan\theta} x\tan\theta - \frac{g}{2v^2\cos^2\theta}x^2 \ dx$$

which is a simple integral (polynomial in $x$).


It is well known, and relatively easy to prove using calculus, that the area under a parabolic arch is $A=\dfrac{2}{3}bh$ where $b$ is the base of the arch and $h$ its height.

The height of a parabolic arch $f(x)=ax^2+bx+c$ where $a,b,c$ make sense is $f(t)$ where $t=-\dfrac{b}{2a}$. In your case $t = \dfrac{v^2\cos^2\theta\tan\theta}{g}$, so

$$ h=\tan\theta\left(\dfrac{v^2\cos^2\theta\tan\theta}{g}\right) - \dfrac{g}{2v^2\cos^2\theta}\left(\dfrac{v^2\cos^2\theta\tan\theta}{g}\right)^2$$

And the base $b$ is the distance between the roots,


This could be a useful exercise in algebra/identities if nothing else.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.