Solutions of $0 = A\sin^2(\alpha) + B\sin(2\alpha) - C$?

Question

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?

Answer 1

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 $$ \tan\alpha=\frac{-B\pm\sqrt{B^2+d^2}}{d} $$

Answer 2

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

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

Answer 3

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 $t$.