# The trajectory of a projectile launched from a hilltop

Here is the problem:

A boy stands at the peak of a hill which slopes downward uniformly at angle $$\phi$$. At what angle $$\theta$$ from the horizontal should he throw a rock so that it has the greatest range?

I realize that the same question is posted here: https://physics.stackexchange.com/questions/24235/trajectory-of-projectile-thrown-downhill, but I have some questions that were not answered in that thread:

1. Can the problem be solved without a rotation of the coordinate system? If so, how?
2. I tried to solve the problem using a rotated coordinate system, but cannot figure out how to finish it (see the work given below).

Here is what I have so far:

1. We set up the coordinate system so that the positive $$x$$ axis coincides with the downward slope of the hill. This simplifies the problem by allowing us to easily relate $$\phi$$ and $$\theta$$, through the relation $$\alpha = \phi + \theta$$.
2. $$v_{0x} = v_0 \cos\alpha$$
3. $$v_{0y} = v_0 \sin\alpha$$
4. $$a_x=-g \cos(\phi-\frac{\pi}{2})=-g \cos (-(\frac{\pi}{2}-\phi))= g\cos(\frac{\pi}{2}-\phi)=g\sin\phi$$
5. $$a_y=-g \sin(\phi-\frac{\pi}{2})=-g \sin (-(\frac{\pi}{2}-\phi))= -g\sin(\frac{\pi}{2}-\phi)=-g\cos\phi$$
6. $$v_x=v_{0x}+\int_{0}^{t}{a_x(t \prime)dt \prime}=v_0 \cos \alpha+\int_{0}^{t}{(g\sin\phi) dt \prime} =v_0 \cos \alpha + t(g\sin\phi)$$
7. $$x=x_0 + \int_0^t{v_x(t\prime) dt\prime} = \int_0^t{(v_0 \cos \alpha + t\prime(g\sin\phi)) dt\prime}=t(v_0 \cos \alpha) + \frac{1}{2}t^2(g \sin \phi)$$
8. $$v_y=v_{0y}+\int_{0}^{t}{a_y(t \prime)dt \prime}=v_0 \sin \alpha+\int_{0}^{t}{(-g\cos\phi) dt \prime} =v_0 \sin \alpha - t(g\cos\phi)$$
9. $$y=y_0 + \int_0^t{v_y(t\prime) dt\prime} = \int_0^t{(v_0 \sin \alpha - t\prime(g\cos\phi)) dt\prime}=t(v_0 \sin \alpha) - \frac{1}{2}t^2(g \cos \phi)$$
10. To find the flight time of the projectile, we find the time at which its trajectory intersects the ground (in this case, the $$x$$ axis), by setting $$y=0$$ and solving for $$t$$. $$y=t(v_0 \sin \alpha) - \frac{1}{2}t^2(g \cos \phi)=0$$ $$v_0 \sin \alpha = \frac{1}{2}t(g \cos \phi)$$ $$t=\frac{2v_0 \sin\alpha}{g \cos \phi}$$
11. Substituting $$t$$ into the equation for $$x$$ gives us the distance traveled by the projectile as a function of the angles $$\alpha$$ and $$\phi$$.$$x=t(v_0 \cos \alpha) + \frac{1}{2}t^2(g \sin \phi)$$ $$x=(\frac{2v_0 \sin\alpha}{g \cos \phi})(v_0 \cos \alpha) + \frac{1}{2}(\frac{2v_0 \sin\alpha}{g \cos \phi})^2(g \sin \phi)$$ $$x=\frac{2v_0^2}{g \cos \phi}(\sin \alpha \cos \alpha)+\frac{2v_0^2}{g \cos \phi}(\sin^2\alpha \frac{\sin\phi}{\cos\phi})$$ $$x=\frac{2v_0^2}{g \cos \phi}(\sin\alpha\cos\alpha+\sin^2\alpha\tan\phi)$$
12. I noticed that the solution in the other thread proceeds from here by differentiating $$x$$ with respect to $$\alpha$$, holding $$\phi$$ constant, which gives $$\frac{dx}{d\alpha}=\frac{2v_0^2}{g \cos \phi}(\frac{d}{d\alpha}(\frac{1}{2}(\sin(2\alpha)+\sin^2\alpha\tan\phi))$$ $$\frac{dx}{d\alpha}=\frac{2v_0^2}{g \cos \phi}(\cos(2\alpha)+2\sin\alpha\cos\alpha\tan\phi)$$ $$\frac{dx}{d\alpha}=\frac{2v_0^2}{g \cos \phi}(\cos(2\alpha)+\sin(2\alpha)\tan\phi)$$ This equation allows us to examine how $$x$$ changes with respect to $$\alpha$$. We see that $$x$$ increases as $$\alpha$$ increases, up to a certain point, and then decreases as $$\alpha$$ increases past this value. This means that the graph of $$x$$ has a relative maximum at the value of $$\alpha$$ which produces the maximum range.
13. We want to find the value of $$\alpha$$ that results in the maximum range of the projectile. In other words, we must determine the value of $$\alpha$$ for which the graph of $$x$$ has a relative maximum. We achieve this by setting $$\frac{dx}{d\alpha}=0=\frac{2v_0^2}{g \cos \phi}(\cos(2\alpha)+\sin(2\alpha)\tan\phi)$$ Dividing each side by $$\frac{2v_0^2}{g \cos \phi}$$ produces $$\cos(2\alpha)+\sin(2\alpha)\tan\phi=0$$ Here is where I get lost. It seems like this should be the easy part, because the only thing left to do is solve the above equation for $$\alpha$$, but I don't know how to do it. Could someone please explain this part to me?

Additionally, I would like to know whether the problem can be solved without rotating the coordinate system. I originally set out to solve it using the standard rectangular coordinate system, but got bogged down in some equations that seemed to lead nowhere. Thanks for your help.

• Rectangular coordinates: $y$, as positive height, and $x$, as positive distance, are given as functions of time. You can equate these two through the line $y=-x\tan(\phi)$. Then the horizontal range becomes $\frac{v^2}{g}(\sin(2\theta) + \tan(\phi)\cos(2\theta) + \tan(\phi))$ which, after maximizing, leads you somewhere close to where you were.
– gmz
Sep 29, 2021 at 22:58

I didn't go through all of your work but to answer the question about how to solve that equation, change $\tan(\phi)$ to $\frac{\sin(\phi)}{\cos(\phi)}$ and take $\cos(\phi)$ as a common denominator. You get $$\frac{\cos(2\alpha)\cos(\phi) + \sin(2\alpha)\sin(\phi)}{\cos(\phi)} = 0$$
Now apply the trig identity $\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$ and your equation reduces to $$\cos(\phi - 2\alpha) = 0$$.