Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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: http://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.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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 $$\cos(\phi - 2\alpha) = 0$$.

share|improve this answer
    
Thanks for clarifying that. It seems obvious now, but earlier I didn't realize that I should put everything in terms of sine and cosine functions in order to solve for the angle. –  fctaylor25 Jul 25 '12 at 13:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.