# Homework Help: Throwing a rock up a hill

1. Mar 8, 2014

### Bryson

1. The problem statement, all variables and given/known data

I was tutoring the other day, when we came across a problem that had me stumped!

A person standing on a hill that forms an angle $\theta = 30^o$ wrt to the horizon, throws a stone at ${\bf v} = 16$ m/s up the hill at an angle $\phi = 65^o$ wrt to the horizon. Find $y_f$.

2. Relevant equations

$$y_f = v_{y_i}t + \frac{1}{2}a_yt^2$$

$$v_{x_i} = \frac{x_f}{t}$$

3. The attempt at a solution

My thought process is the first find time (t), then solve for $y_f$.

Initially I thought to take the ratio of $v_x$ and $v_y$, which would result in an equation involving $tan(\phi - \theta)$, but it involves too many unknowns (t, $x_f$).

I know I need to utilize the angles in someway, and that finding $h_{max}, R$ will not help in this situation. Any suggestions on how to start this problem would be greatly appreciated!

2. Mar 8, 2014

### Staff: Mentor

Hint: Find an expression for y as a function of x.

3. Mar 8, 2014

### Bryson

I tried this but here is what I got: $$y_f = x_f tan(\phi - \theta) + \frac{1}{2}a_y t^2.$$ Still too many unknowns though, assuming my algebra and logic is correct of course.

4. Mar 8, 2014

### Staff: Mentor

Get rid of time in that equation. Not sure why you are subtracting angles at this point.

Find the trajectory of the rock, then compare that to an equation describing the hill.

5. Mar 8, 2014

### CWatters

Getting rid of t is usually straight forward. You know the horizontal velocity and can write an equation for the horizontal distance Xf in terms of Yf from the slope.

6. Mar 8, 2014

### Bryson

Getting rid of time; the only way I see that is if I substitute $t = \frac{x_f}{v_x}$, giving us

$$y_f = x_f tan(\theta) + \frac{1}{2 v_x^2} a_y x_f^2.$$

Perhaps I do not understand, we do not know $y_f$, nor $x_f$.

7. Mar 8, 2014

### Bryson

Yep, got it. I was on the wrong track. . . I would shoot myself in the foot, but instead I may get rid of it altogether!

Thanks for the help and quick replies.