1. Peak of a hill

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 is has the greatest range.

Ok, so this is a rotation of the normal $x_{1} - x_{2}$ plane right? So we can use the direction cosines $\lambda_{ij}$ to make this problem easier.

So $x'_{1} = x_{1} \cos \phi + x_{2} \cos \left(\frac{\pi}{2} + \phi \right)$ and $x'_{2} = x_{1}\cos \theta + x_{2}\cos \phi$.

Are these the right transformations? Is this the right way to set up the problem?

2. I am using the following:

$\begin{bmatrix} x'_{1} \\ x'_{2} \\ x'_{3} \end{bmatrix} = \begin{bmatrix} \lambda_{11} & \lambda_{12} & \lambda_{13} \\ \lambda_{21} & \lambda_{22} & \lambda_{23} \\ \lambda_{31} & \lambda_{32} & \lambda_{33} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$

Then $\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} \lambda_{11} & \lambda_{12} & \lambda_{13} \\ \lambda_{21} & \lambda_{22} & \lambda_{23} \\ \lambda_{31} & \lambda_{32} & \lambda_{33} \end{bmatrix}^{T} \begin{bmatrix} x'_1 \\ x'_2 \\ x'_3 \end{bmatrix}$

3. Originally Posted by shilz222
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 is has the greatest range.

Ok, so this is a rotation of the normal $x_{1} - x_{2}$ plane right? So we can use the direction cosines $\lambda_{ij}$ to make this problem easier.

So $x'_{1} = x_{1} \cos \phi + x_{2} \cos \left(\frac{\pi}{2} + \phi \right)$ and $x'_{2} = \cos \theta + \cos \phi$.

Are these the right transformations? Is this the right way to set up the problem?
I wouldn't bother with direction cosines. If you rotate the problem then you have two acceleration components to deal with, along with sacrificing any physical intuition you might have developed.

I am going to set up a coordinate system with the origin at the peak (an unusual placement for me), with a +y direction straight up, and a +x direction directly to the right. (The hill in my diagram slopes down and to the right.)

I am going to call the "angle of inclination" (of the initial velocity) $\theta$. Note that this angle could actually be below the +x axis, though my intuition right now says that it won't be. But we should keep the idea in mind, just to be safe. Additionally the rock will be thrown at an initial speed $v_0$. We aren't given this and I will presume that it either drops out or won't be required.

So we know that the rock has the following information initially:
$x_0 = 0~m~~~~~y_0 = 0~m$
$v_{0x} = v_0~cos(\theta)~~~v_{0y} = v_0~sin(\theta)$
$a_x = 0~m/s^2~~~~a_y = -g$
where $g = 9.8~m/s^2$.

For the information at the point of impact we have
$t = ~?$ <-- Time of impact
$x = ~?~~~~~y = ~?$
$v_x = v_{0x}~~~~v_y = ~?$
( $v_x = v_{0x}$ is known because $a_x = 0~m/s^2$.)

We also know that $\frac{-y}{x} = tan(\phi)$. (The y value at impact must be negative.)

So. How to proceed? Well, we know that
$x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

$x = v_0t~cos(\theta)$

We also know
$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

$y = v_0t~sin(\theta) - 4.9t^2$

So
$\frac{-y}{x} = \frac{4.9t^2 - v_0t~sin(\theta)}{v_0t~cos(\theta)} = tan(\phi)$

$\frac{4.9t - v_0~sin(\theta)}{v_0~cos(\theta)} = tan(\phi)$

So
$t = \frac{v_0~cos(\theta)~tan(\phi) + v_0~sin(\theta)}{4.9}$

(Where am I going with this? Keep your eye on the goal: we want the maximum range, so we need an expression for x in terms of $\theta$.)

Now plug this into the x equation:
$x = v_0t~cos(\theta)$

$x = v_0 \left ( \frac{v_0~cos(\theta)~tan(\phi) + v_0~sin(\theta)}{4.9} \right ) ~cos(\theta)$

$x = \frac{v_0^2~tan(\phi)}{4.9} \cdot cos^2(\theta) + \frac{v_0^2}{4.9} \cdot sin(\theta)~cos(\theta)$

Critical points for the function $x(\theta)$ can be found by:
$\frac{dx}{d \theta} = \frac{v_0^2~tan(\phi)}{4.9} \cdot 2 cos(\theta) \cdot -sin(\theta) + \frac{v_0^2}{2 \cdot 4.9} \cdot 2cos(2 \theta)$
(I cheated and used $sin(\theta)~cos(\theta) = \frac{1}{2}sin(2 \theta)$.)

So the critical points are at
$\frac{v_0^2~tan(\phi)}{4.9} \cdot 2 cos(\theta) \cdot -sin(\theta) + \frac{v_0^2}{4.9} \cdot cos(2 \theta) = 0$

$-\frac{v_0^2~tan(\phi)}{4.9} \cdot sin(2 \theta) + \frac{v_0^2}{4.9} \cdot cos(2 \theta) = 0$

$\frac{v_0^2~tan(\phi)}{4.9} \cdot sin(2 \theta) = \frac{v_0^2}{4.9} \cdot cos(2 \theta)$

$tan(2 \theta) = \frac{ \frac{v_0^2}{4.9} }{ \frac{v_0^2~tan(\phi)}{4.9} }$

$tan(2 \theta) = \frac{1}{tan(\phi)}$

So we have the result that
$\theta = \frac{1}{2} atn(cot(\phi))$
gives the angle for the initial velocity to give the maximum range. (Whew! You can prove that this is a relative maximum and not a relative minimum.)

My cluttered brain is thinking there's a way to simplify this, but I'm just not sure. My thought is that
$\theta = \frac{1}{2} atn(cot(\phi)) = \frac{1}{2} \left ( \frac{\pi}{2} - \phi \right )~rad$
but don't quote me on this.

-Dan

4. hint is given: $\phi = 60, \ \theta = 15$.

5. Originally Posted by shilz222
hint is given: $\phi = 60, \ \theta = 15$.
Okay, so my simplification works for two cases: the one you just mentioned and for $\phi = 0$ in which case the maximum range is provided by $\theta = \frac{\pi}{4}$. (A much simpler problem to work out!)

-Dan