Throwing a Rock Up a Hill: Finding Final Vertical Displacement

[Bryson]

Homework Statement

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.

Homework Equations

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

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

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!

 

[Doc Al]

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

 

[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.

 

[Doc Al]

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.

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.

 

[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.

 

[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.

 

[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.