# Newtonian mechanics - hillside fired projectile

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1. Oct 5, 2015

### KUphysstudent

1. The problem statement, all variables and given/known data
A canon is burried on a hillside, so the barrel is exactly at the height of the surface. The hillside creates angle theta with vertical, where 0 degrees < theta < 90 degrees. A canonball is fired with speed v_0 perpendicular on the hillside. The angle between velocity_0 and horizontal also has angle theta. g is 9.8 and there is no air resistance.

a) Find the vertical distance from the firingpoint to the point where the canonball again has the same height as when it was fired.

b) Find the horizontal distance from the firingpoint to the impact location on the hillside.

This is the exact text translated from my language. And sorry about my lack of paint skills.
2. Relevant equations
How am I supposed to find the distance when I have no variables?

3. The attempt at a solution
vox = v0*cosθ
voy = v0*sinθ

y = voy*t-1/2*g*t2
y = 0 = voy*t2-1/2*g*t2 = t2(voy-1/2*g*t2)
t2=0 and t2=(2*voy)/g
R = vox*t2

This is my first time using this forum, so I hope this is the style people want to see it in.

2. Oct 5, 2015

### RUber

I have a couple of questions,
1) Does vertical distance imply total distance travelled between initial point and apex and back again--i.e. twice the max height above firing point?
2) Does impact location on the hillside imply that the projectile is impacting the same hillside, with slope of -cot(theta)?
In answer to your question--without additional information, you will be looking for solutions in terms of v_0 and theta.

3. Oct 5, 2015

### HallsofIvy

You are told that the hillside makes angle $\theta$ with the vertical which means it has angle, as in your picture, $-\theta$ (radians) with the horizontal. If we take the position of the barrel of the cannon to be (0, 0) in an xy- coordinate system with the y-axis vertical and x-axis horizontal we can write the equation of the line, as shown in your picture, $y= -tan(\theta)x$. The cannon ball is fired perpendicular to the hillside so at angle $\theta$ to the horizontal, as you show. The x and y components are as you give, $y(t)= -(g/2)t^2+ v_0 sin(\theta)t$ and $x(t)= v_0 cos(\theta)$. You want to find the point where the two graphs intersect. Replace the x and y in $y= -tan(\theta)x$ with $y(t)= -(g/2)t^2+ v_0 sin(\theta)t$ and $x(t)= v_0 cos(\theta)$ to get a single quadratic equation in t. Of course, one of the two solutions will be t= 0.

4. Oct 5, 2015

### KUphysstudent

1] vertical distance is from the initial point and to the point where it impacts the x-axis again. So a normal parabola with initial velocity and angle. that is what question (a) is about.
2] Then question (b) is where the projectile impacts the hillside when it goes under the x-axis.
Im sorry if my question is confusing, but I just translated the text in the assignment.
Also I have changed the picture to add the two parabolas hopefully helping, though they were not on the picture giving with the assignment.

b

By the way thanks for your quick answer. myself is often that I dont know if you in some way to calculate the result in some sneaky way or they just want the formulars.

5. Oct 5, 2015

### RUber

In that case, it looks like you have part (a) solved.
For part (b), define your points on the hill as pairs (x, f(x)) and do the same for your position function. I mean change t into a function of x, and use that in your functional form for y(t) to make y(x).
Then you should be able to 'generally' solve for where the curves intersect.

6. Oct 5, 2015

### KUphysstudent

½ * g * t2 - (v0 * sinθ) * t + y = 0
t = (v0 * sinθ ± sqrt((-v0 * sinθ) - 4 (½ * g) * y)) / 2(½ * g)
t = (v0 * sinθ ± sqrt(v02 * sin2θ - 2 * g * y)) / g
Using the positive root
x = (v0 * cosθ) * t

I did it like this earlier, I just thought it wasnt good enough because I couldnt get a numerical result.