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I am trying to solve a task in which I need to calculate the optimal angle($\alpha$) with which the projectile will land the furthest from a height ($h$), so what I basically have is the equations for the projectile's movement:

$$ x = v_0 \cos\alpha t $$ $$ y = h + v_0\ \sin\alpha t - \frac{gt^2}{2} $$ $$ t_{y=0} = \frac{v_0\sin\alpha+\sqrt{(v_0\sin\alpha)^2 + 2gh}}{g} $$ I know that at height $h=0$ the optimal angle is $\alpha = 45^\circ$ and I can calculate the range $x$, but I don't know how to calculate the angle.

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closed as off-topic by Brandon Enright, user10851, Kyle Kanos, Dan, David Z Jan 12 '14 at 6:57

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    $\begingroup$ Lateral comment: perhaps a slightly easiest way than passing through time to find $x_{\text {range}}(\alpha)$ is to substitute $t=x/(v_0 \cos \alpha)$ in $y$'s equation and solving $y=0$. $\endgroup$ – pppqqq Jan 12 '14 at 10:04
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It will help to scale the problem by defining the dimensionless variable $$ \delta=\frac{2gh}{v_0^2}. $$ Notice that we can take the limit $\delta\rightarrow0$ to recover the solution when $h=0$ in which case we expect to get $\alpha=\tfrac{\pi}{4}$. With this substitution your expression for the time at which the projectile hits the ground becomes $$ t_{y=0}=\frac{v_0}{g}\left(\sqrt{\sin^2\alpha+\delta}+\sin\alpha\right). $$ Substituting this into the expression for $x(t)$ gives $$ x(t_{y=0})=\frac{v_0^2}{g}\cos\alpha\left(\sqrt{\sin^2\alpha+\delta}+\sin\alpha\right). $$ It is this expression that we want to maximize with respect to $\alpha$. Taking the first derivative gives $$ \frac{d}{d\alpha}x(t_{y=0})= \frac{v_0^2}{g}\left(\cos^2\alpha\left(\frac{\sin\alpha} {\sqrt{\sin^2\alpha+\delta}}+1\right) -\sin\alpha\left(\sqrt{\sin^2\alpha+\delta}+\sin\alpha\right) \right). $$ Now we set this equal to zero and solve for $\alpha$. At the outset, I honestly didn't think the problem would have a closed form solution, but Mathematica had no trouble inverting this equation. The final result (choosing the physical result) is $$ \alpha=\arccos\left(\frac{\sqrt{\delta+1}}{\sqrt{\delta+2}}\right) $$ We can check this solution at the usual limits; at $\delta=0$ we get $\alpha=\tfrac{\pi}{4}$ and at $\delta=\infty$ (the platform is very high) we get $\alpha=0$ both of which sound good. Notice that for $\delta<-1$ the solution gives imaginary results which are unphysical. This is because when $\delta<-1$ the launch platform is so far underground that the initial speed $v_0$ is not even enough to get it to the surface. With this in mind, we can check one final limit of the problem; if $\delta=-1$, then the initial velocity is just enough to get the projectile to $y=0$ and the launch angle we find is $\alpha=\tfrac{\pi}{2}$ which is pointing the gun straight up.

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Substitute $t$ into $x(t)$ to get the throw distance and maximize this expression with respect to angle $\alpha$, i.e. set its derivative zero and solve for optimal the angle.

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