Projectile fired on hill at angle

In summary: If you think about it, the easiest way to do it would be to use a graphing calculator to plot the functions and then find the intersection.
  • #1
eku_girl83
89
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!
 
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  • #2
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.
 
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  • #3
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...

[tex] \tan\beta = \frac{x}{y} [/tex]

[tex]x = V_{xo}t = \cos({\alpha})V_{o}t[/tex]

[tex]y = V_{yo}t + \frac{1}{2}gt^2 = \sin({\alpha})V_{o}t + \frac{1}{2}gt^2
[/tex]
 
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  • #4
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.
 

1. What is a projectile fired on a hill at an angle?

A projectile fired on a hill at an angle refers to an object that is launched from a point on a hill and follows a curved path due to the combined effects of its initial velocity and the force of gravity.

2. How is the angle of launch determined for a projectile fired on a hill?

The angle of launch for a projectile fired on a hill is determined using trigonometry, specifically the tangent function. The angle is typically measured from the horizontal plane.

3. What factors affect the trajectory of a projectile fired on a hill at an angle?

The trajectory of a projectile fired on a hill at an angle is affected by several factors, including the initial velocity, angle of launch, gravitational force, air resistance, and the shape and mass of the projectile.

4. How does the height of the hill affect the flight of a projectile?

The height of the hill can affect the flight of a projectile by changing the angle of launch and therefore altering the initial velocity and trajectory. A higher hill may also result in a longer flight time and greater horizontal displacement.

5. What is the maximum range for a projectile fired on a hill at an angle?

The maximum range for a projectile fired on a hill at an angle is achieved when the angle of launch is 45 degrees. This results in the greatest horizontal displacement and is known as the optimum angle of launch.

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