Projectile fired on hill at angle

AI Thread Summary
The discussion revolves around calculating the trajectory of a projectile fired at an angle up a hill with a slope. Participants express confusion about deriving the equations of motion due to the complexity of firing at an angle while considering the hill's slope. Key points include the need to break down the projectile's motion into components and the suggestion to find the intersection of the projectile's path and the hill's equation. There is also mention of using numerical methods if specific values are provided, but the goal is to derive an algebraic solution. The conversation highlights the challenges of maximizing the range and determining the landing point on the hill.
eku_girl83
Messages
89
Reaction score
0
This one is giving me fits!

A projectile is fired with initial speed v0 at an elevation angle of alpha up a hill of slope beta (alpha greater than beta).

How far up the hill will the projectile land?
At waht angle alpha will the range be a maximum?
What is the maximum range?

Where do I begin? How do I write (or alternately, derive) the equations of motions for this case? Normally, for a projectile fired at an angle, we divide the velocity into components and then apply equations for two-dimensional motion. But I just don't know how to approach a projectile at an angle which is fired at another angle!

Any help greatly appreciated!
 
Physics news on Phys.org
When I read it, I was thinking the place from where you fire the projectile from is flat and then there's a hill in front of it. (ie alpha and beta both measured from the x-axis)

Sounds like you're saying alpha is the angle between the projectile and the hill, if that's the case then you just add the two angles to get the angle at which the projectile is fired (relative to the x-axis). In that case though, why would it say "alpha greater than beta".

Either way you choose to use the angle... you can still get your *time when ball hits hill* using the Y motion of the projectile, you'll just have an extra term in there that'll relate how the height of the ground changes over time.
 
Last edited:
That does seem like a tough one. I think you could do it numerically if you were given values but it sounds like you are wanted to do it algebracially. I'm a bit bit stumped by it myself, this might help but probably not :confused: . If I have the problem set up right, I believe you can set it up for the tan(beta), then have that equal to the equation for x over the equation for y. I'm not sure what to do to maximize x for alpha though. Ok, hope I can get this TeX right...

\tan\beta = \frac{x}{y}

x = V_{xo}t = \cos({\alpha})V_{o}t

y = V_{yo}t + \frac{1}{2}gt^2 = \sin({\alpha})V_{o}t + \frac{1}{2}gt^2<br />
 
Last edited:
I believe this situation boils down to finding the intersection of 2 functions. One function is the y = f(x) to describe the projectile as if there were no incline. The second function describes the incline, which is just a line. Find the intersection of the two functions.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top