MHB How can you find the best angle and range for a projectile shot from a building?

[Euge]

Here's this week's problem!

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Problem. A projectile is shot from the edge of a building of height $h$ with initial speed $v$ at an angle $\alpha$ that gives the greatest range $d$. Show that $$\alpha = \cos^{-1}\left(\sqrt{\frac{2gh + v^2}{2gh + 2v^2}}\right) \quad \text{and} \quad d = \frac{v}{g}\sqrt{2gh + v^2}.$$
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[Euge]

This week's problem was correctly solved by Rido12. You can find his solution below.

Spoiler

Assuming that air friction is negligible and the acceleration is constant, $g$. Thus, we have the following two equations:
$$x=v_xt+\frac{1}{2}a_xt^2+x_0=v_xt $$
$$y=v_yt+\frac{1}{2}a_yt^2+h$$

Now, the projectile will reach the ground at $y=0$, so solving the second equation for t, and taking the positive root, we obtain: (where $g=-a$)
$$t=\frac{v\sin(\alpha)+\sqrt{v^2\sin(\alpha)+2gh}}{g}$$
Re-substituting $t$ into the first equation:

$$R=x=v\cos(\alpha)\left(\frac{v\sin\left({\alpha}\right)+\sqrt{v^2\sin^2\left({\alpha}\right)+2gh}}{g}\right)$$

Differentiating $R$, the range with respect to $\alpha$, we obtain:
$$\d{R}{\alpha}=\frac{v^2\sin\left({\alpha}\right)\cos^2\left({\alpha}\right)}{\sqrt{v^2\sin^2\left({\alpha}\right)+2gh}}+v\left(\cos^2\left({\alpha}\right)-\sin^2\left({\alpha}\right)\right)-\sin\left({\alpha}\right)\sqrt{v^2\sin^2\left({\alpha}\right)+2gh}$$

Simplifying and setting equal to $0$:

$$\sin(\alpha)=\frac{v}{\sqrt{2v^2+2gh}}$$
$$\cos(\alpha)=\cos\left({\sin^{-1}\left({\frac{v}{\sqrt{2v^2+2gh}}}\right)}\right)=\sqrt{\frac{2gh+v^2}{2gh+2v^2}}$$
$$\alpha=\cos^{-1}\left({\sqrt{\frac{2gh+v^2}{2gh+2v^2}}}\right)$$

Now, re-substituting into the range equation $R=v\cos(\alpha)\left(\frac{v\sin\left({\alpha}\right)+\sqrt{v^2\sin^2\left({\alpha}\right)+2gh}}{g}\right)$, where $\sin(\alpha)=\frac{v}{\sqrt{2v^2+2gh}}$ and $\cos(\alpha)=\sqrt{\frac{2gh+v^2}{2gh+2v^2}}$ : (the algebra is tedious and has therefore been omitted)
$$d = \frac{v}{g}\sqrt{2gh + v^2}$$
as required.

Even though the general method behind this solution is sound, the missing algebra is nontrivial and very crucial to this problem. I'll show a method of finding $\alpha$ below:

Spoiler

I'll start with the expression

$$\frac{dR}{d\alpha} = \frac{v^2}{g}\left\{\cos^2 \alpha + \frac{\sin \alpha \cos^2 \alpha}{\sqrt{\sin^2 \alpha + \frac{2gh}{v^2}}} - \sin^2 \alpha - \sin \alpha \sqrt{\sin^2 \alpha + \frac{2gh}{v^2}}\right\}.$$

Setting $\frac{dR}{d\alpha} = 0$, and letting $\ell = 2gh/v^2$ and $u = \cos^2\alpha$, we have

$$u + u\sqrt{\frac{1 - u}{1 - u + \ell}} - (1 - u) - \sqrt{1 - u} \sqrt{1 - u + \ell} = 0$$

$$(2u - 1) + \sqrt{1 - u}\left(\frac{u}{\sqrt{1 - u + \ell}} - \sqrt{1 - u + \ell}\right) = 0$$

$$(2u - 1) + \sqrt{1 - u} \left(\frac{u - (1 - u + \ell)}{\sqrt{1 - u + \ell}}\right) = 0$$

$$(2u - 1) + \sqrt{1 - u} \left(\frac{2u - 1 - \ell}{\sqrt{1 - u + \ell}}\right) = 0$$

$$\frac{1 - u}{1 - u + \ell} (2u - 1 - \ell)^2 = (1 - 2u)^2$$

$$\frac{1 - u}{1 - u + \ell}[(2u - 1)^2 - 2\ell(2u - 1) + \ell^2] = (2u - 1)^2 (1 - u + \ell)$$

$$(1 - u)[\ell^2 - 2\ell(2u - 1)] = (2u - 1)^2 \ell$$

$$\ell(1 - u)[\ell - 2(2u - 1)] = (2u - 1)^2\ell$$

$$(1 - u)[(2u - 1)^2 - 2\ell(2u - 1) + \ell^2] = (1 - 2u)^2 (1 - u + \ell)$$

$$(1 - u)[\ell^2 - 2\ell(2u - 1)] = (2u - 1)^2\ell$$

$$\ell^2(1 - u) - 2\ell(2u - 1)(1 - u) = (2u - 1)^2 \ell$$

$$\ell^2 - \ell^2 u - 2\ell[2u - 1 - 2u^2 + u] = (4u^2 - 4u + 1)\ell$$

$$\ell^2 - \ell^2 u - 4\ell u + 2\ell + 4u^2\ell - 2\ell u = 4u^2 \ell - 4u\ell + \ell$$

$$\ell^2 - \ell^2 u + \ell - 2\ell u = 0$$

$$\ell - \ell u + 1 - 2u = 0$$

$$(\ell + 1) - (\ell + 2)u = 0$$

$$u = \frac{\ell + 1}{\ell + 2}$$

$$\cos^2 \alpha = \frac{\ell + 1}{\ell + 2}$$

$$\cos \alpha = \sqrt{\frac{\ell + 1}{\ell + 2}}$$

$$\cos \alpha = \sqrt{\frac{\frac{2gh}{v^2} + 1}{\frac{2gh}{v^2} + 2}}$$

$$\cos \alpha = \sqrt{\frac{2gh + v^2}{2gh + 2v^2}}$$

$$\alpha = \cos^{-1} \sqrt{\frac{2gh + v^2}{2gh + 2v^2}}$$