2
$\begingroup$

I am facing a problem from physics class involving a projectile motion which can be described with such an given equation: $$ h = -\frac{1}{2} \frac{g}{v_{0}^2 \cos^2\alpha} d^2 + \frac{\sin\alpha}{\cos\alpha} d + y. $$ The goal is to find the minimum value of (rearranging above equation) $$ v_0(\alpha) = \frac{d}{\cos\alpha} \cdot \sqrt{\frac{1}{2} \cdot \frac{g}{\tan\alpha \cdot d + y - h}}. $$ This involves finding solutions to $v_0'(\alpha) = 0$. I was able to find the derivative ($t := d\cdot \tan\alpha +y-h$): $$ v_0'(\alpha) = \frac{\tan \alpha}{\cos\alpha \cdot \sqrt{t}} - \frac{d}{2\cos^3\alpha \cdot (\sqrt{t})^3} $$ Because of the condition $v_0'(\alpha) = 0$ this simplifies to (it is know that the solution is around $50^\circ$) $$ 0 = \sin\alpha - \frac{d}{2\sin\alpha \cos\alpha \cdot (d\cdot \tan\alpha + y - h)} $$ or $$ 0 = d\sin(2\alpha)\tan\alpha + (y-h)\sin(2\alpha) - d. $$ or $$ 0 = 2d\sin^2(\alpha) + (y-h)\sin(2\alpha) - d $$ or with some constants

$$ 0 = 2d\sin^2(\alpha) + B\sin(2\alpha) - d. $$ How can I find the solutions from here?

$\endgroup$
3
  • $\begingroup$ i would use the tan half angle substitution $\endgroup$ – Dr. Sonnhard Graubner Nov 2 '17 at 20:21
  • $\begingroup$ How does the identity look? $\endgroup$ – Simon Mueller Nov 2 '17 at 20:23
  • $\begingroup$ Note that $2\sin^2\alpha-1 = -\cos 2\alpha$; this allows you to find $\tan 2\alpha$ in a pretty simple form. $\endgroup$ – rogerl Nov 2 '17 at 20:52
1
$\begingroup$

You can expand $\sin2\alpha$: $$ 2d\sin^2\alpha+2B\sin\alpha\cos\alpha-d\sin^2\alpha-d\cos^2\alpha=0 $$ that becomes $$ d\sin^2\alpha+2B\sin\alpha\cos\alpha-d\cos^2\alpha=0 $$ If $d\ne0$, we cannot have $\cos\alpha=0$ as a solution, so we can divide by $\cos^2\alpha$, leading to $$ d\tan^2\alpha+2B\tan\alpha-d=0 $$ so $$ \tan\alpha=\frac{-B\pm\sqrt{B^2+d^2}}{d} $$

$\endgroup$
0
$\begingroup$

Are you not comfortable with the good old $$ y = \tan\alpha \cdot x-\frac{1}{2} \frac{g}{v_{0}^2 \cos^2\alpha} x^2 $$

expression of the projectile graph that passes through $(0,0)?$

You added, as an initial condition $y$, which is normally reseved for variables.It should be clear what is a constant and what is a variable.

And more directly you can obtain the maximum altitude by setting $ \dot y=0 $ or $ \dfrac{dy}{dx} =0 $

That is directly by uniform deceleration maximum height is

$$ h=\dfrac {{(v_0 \sin \alpha)}^2}{2 g}, v_0 = \dfrac{ \sqrt{2gh}}{\sin \alpha}. $$

$\endgroup$
1
  • $\begingroup$ The OP is trying to find the minimum initial velocity that can achieve a given maximum height. $\endgroup$ – rogerl Nov 2 '17 at 21:01
0
$\begingroup$

Let $\sin^2\alpha=t$ and write

$$At\pm2B\sqrt{t(1-t)}=C,$$ then

$$4B^2t(1-t)=(C-At)^2.$$

Solve the quadratic equation for $t$.

$\endgroup$

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.