Projectile on an Incline

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  • #1
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Homework Statement:
A very tall hill has a steady incline 9 degrees above the horizontal. A cannon is at the bottom of the hill and aimed an angle 25 degrees above the incline of the hill. After the cannon s fired the ball lands 55 meters up the hill (measured along the incline). Calculate the launch speed of the cannon ball.
Relevant Equations:
Basic Kinematic Equations
I have tried for a couple hours now and I just cannot seem to figure out what to do. I am able to to find the the x and y components for acceleration (Xa = 0, Ya = -9.8) and distance (Δx = 54.32, Δy = 8.60) but without any of the other components I am having trouble finding Vo. Any help with pointing me in the right direction would be greatly appreciated.
 

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  • #2
PhanthomJay
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Problem Statement: A very tall hill has a steady incline 9 degrees above the horizontal. A cannon is at the bottom of the hill and aimed an angle 25 degrees above the incline of the hill. After the cannon s fired the ball lands 55 meters up the hill (measured along the incline). Calculate the launch speed of the cannon ball.
Relevant Equations: Basic Kinematic Equations

I have tried for a couple hours now and I just cannot seem to figure out what to do. I am able to to find the the x and y components for acceleration (Xa = 0, Ya = -9.8) and distance (Δx = 54.32, Δy = 8.60) but without any of the other components I am having trouble finding Vo. Any help with pointing me in the right direction would be greatly appreciated.
You might want to look in both directions and solve 2 simultaneous equations with 2 unknowns.
 
  • #3
gneill
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Start with the basic equations of motion for the cannon ball (x and y components of the motion in terms of the launch angle and time). Note that these are parametric equations in terms of t. Then see if you can't render these into a single equation for the trajectory, i.e. y = f(x). The initial launch speed remains an unknown at this point. This trajectory equation intersects the line that represents the hill's surface at a known location...
 
  • #4
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Ok so after thinking about the feedback I was able to determine that the time = 3.05 seconds. With this I was able to calculate the Initial velocity in all directions. I got the following results:

Vo = 32.98 m/s
Vox = 17.80 m/s
Voy = 17.76 m/s

The problem is that the velocities do no match up using the Pythagorean theorem. I was able to get the time by making Vox = 54.32/t and Voy = 4.9t + (8.6/t) and setting them equal to one another. I guess I'm just confused where I went wrong. Because if I can find Vo at the 9 degree angle I will be able to find it at the 25 degree angle using trigonometry.
 
  • #5
PhanthomJay
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Ok so after thinking about the feedback I was able to determine that the time = 3.05 seconds. With this I was able to calculate the Initial velocity in all directions. I got the following results:

Vo = 32.98 m/s
Vox = 17.80 m/s
Voy = 17.76 m/s

The problem is that the velocities do no match up using the Pythagorean theorem. I was able to get the time by making Vox = 54.32/t and Voy = 4.9t + (8.6/t) and setting them equal to one another. I guess I'm just confused where I went wrong. Because if I can find Vo at the 9 degree angle I will be able to find it at the 25 degree angle using trigonometry.
But Voy and Vox are not equal. They are trig related by vector components at the 25 degree initial angle.
 
  • #6
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Okay so I changed my time equation to 54.32tan(9) = 4.9t^2 + 8.6 and now the Pythagorean theorem works for all the Vo vectors. I got Vo = 2075.07 m/s.

To find Vo at the 25 degree angle I used the Vo I just found as the new x-axis and calculated Vo = 2075.07/cos(25) = 2289.53 m/s.

However this answer seems extremely fast, so I do not think this is correct.
 
  • #7
PhanthomJay
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Okay so I changed my time equation to 54.32tan(9) = 4.9t^2 + 8.6 and now the Pythagorean theorem works for all the Vo vectors. I got Vo = 2075.07 m/s.

To find Vo at the 25 degree angle I used the Vo I just found as the new x-axis and calculated Vo = 2075.07/cos(25) = 2289.53 m/s.

However this answer seems extremely fast, so I do not think this is correct.
Yeah that’s real fast , good observation.

You are getting mixed up on a couple of points. The 9 degree incline has nothing to do with the initial velocity components. It just establishes the x and y coordinates at the landing. It is the 25 degree angle that determines the initial velocity x and y components.

Also, don’t get careless with your equations. The second one is not correct
 
  • #8
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Okay so I realize now that the angle I should be using is 34 degrees because that is where the ball is launched relative the the horizon. With this realization and a reworking of my time equation I got the following equation:

54.32/t = (4.9t + (8.60/t))/tan(34), where t = 2.39 seconds

With t figured out I find Vox = 54.32/2.39 = 22.73

And with Vox I find Vo = 22.73/cos(34) = 27.41

I believe that 27.41 is the final answer. But I could be wrong yet again.
 
Last edited:
  • #9
PhanthomJay
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Looks good now! I'd round it off to 27 m/s
 
  • #10
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Awesome. Thank you so much!
 

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