at what angle is ymax=range?

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anonymous · Answer

I'm assuming you're talking about projectile motion over a flat surface. Let the initial velocity be $v$ and the angle the projectile is fired above the horizontal be $\theta$. The $x$ and $y$ components of the velocity are as follows.$$v_x = v \cos \theta $$$$v_y = v\sin\theta$$We will apply the equation $\Delta \vec r = \vec v_i t + \frac{1}{2}\vec at^2$ to both components recognizing that acceleration vertically is $g$ pointed downward and that there is no horizontal acceleration. Note that the squares of the final velocities in both the x and y directions will be the same by symmetry.$$\Delta x = vt\cos\theta$$$$\Delta y=vt\sin\theta-\frac{1}{2}gt^2$$Now, at the end of the projectile motion when $\Delta t = t_f$, $\Delta x$ here gives us our range while halfway through the projectile motion, when $\Delta t = \frac{t_f}{2}$, $\Delta y$ will give us our range ($t_f$ is the amount of time it takes for the projectile to sweep through the entire arc). Let us substitute these values (we will find the value of $t_f$ later).$$\Delta x = vt_f\cos\theta$$$$\Delta y=\frac{vt_f\sin\theta}{2}-\frac{gt_f^2}{8}$$Now, since $\Delta x=\Delta y$, we can express the following.$$vt_f\cos\theta=\frac{vt_f\sin\theta}{2}-\frac{gt_f^2}{8}$$Now, since $\Delta v = a\Delta t$, we can find $t_f$ by considering the motion along the y-direction through the entire motion. Since the vertical velocity changes from $v\sin\theta$ to $-v\sin\theta$, $\Delta v = -2v\sin\theta$ along with $a=-g$ and $\Delta t = t_f$. $t_f$ is given as follows.$$-2v\sin\theta = -gt_f \Rightarrow t_f = \frac{2v\sin\theta}{g}$$We will now plug this into our equation.$$\frac{2v^2\sin\theta\cos\theta}{g}=\frac{v^2\sin^2 \theta}{g}-\frac{v^2 \sin^2 \theta}{2g}$$We will simplify as well as apply the double-angle formula $2\sin\theta\cos\theta=\sin 2\theta$.$$\sin 2\theta = \frac{1}{2}\sin^2 \theta$$You'd be rather hard-pressed to figure out that intersection on your own, so graphing and finding the intersection is really the best way to go (I used http://www.wolframalpha.com/input/?i=sin+2x+%3D+.5+(sin+x)%5E2 thuogh take note that it ouputs in radians). The solution such that $0\le \theta \le \pi$ are the following.$$\boxed{\theta = \left\{0^\circ, 75.964\ldots^\circ \right\}}$$Yes, in fact $\theta = 0^\circ$ is a solution, as the projectile has no time to leave the ground, so thus both its range and maximum height are both zero (and thus equal). This is called the "trivial solution" (i.e. - obvious solution), and is of lesser signifncance than the other solution. For your edification, the exact value of the nontrivial solution is the following (also found via WolframAlpha). $$\theta = 2\text{arctan}\left(\frac{\sqrt{17}-1}{4}\right)$$-- Finishing remarks: This was a surprisingly challenging problem for its simplicity! I enjoyed it thouroughly! I'm surprised I've never seen this one before.