Projectile Fired Against Sloped Surface

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A projectile fired from a sloped surface requires a different angle for maximum range compared to a flat surface, with the optimal angle depending on the slope's angle. The discussion highlights the challenge of calculating the firing angle (theta) when the launch surface is inclined (phi), noting that traditional methods using kinematics become complex due to trigonometric functions. A hint suggests that for a slope angle of 60 degrees, the optimal firing angle is 15 degrees. One participant ultimately sought assistance but resolved the issue with help from a physics TA, who equated the projectile's height with the slope's height to find the solution. This problem illustrates the complexities of projectile motion on inclined planes.
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Lets say a Projectile is fired at an angle (theta) from the Horizontal. The Catch is that the place it was fired from is a hill sloped angle (phi) against the horizontal. The question is what are the angles are which you can get the highest Range (Range = not the Farthest Horizontal Distance, but the farthest distance along the slop).

Hint: If (phi) is 60, then (theta) is 15.
I've tried doing this problem multiple ways. I set the slope length as D. Making the Vertical displacement vector at the end -Dsign(phi), and the Horizontal Displacement Vector Dcos(phi).

I then used the basic Kinematics equations and combined them. I solved for t under the horizontal displacement and plugged in what I got into my vertical displacement equation, but I keep going nowhere because I have a bunch of sines and cosines with theta and phi.

Can anyone save me?I understand if the ground was flat, the answer would be 45 Degrees as the firing angle, but... with the slope it is VERY different.
 
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CzarValvador said:
Lets say a Projectile is fired at an angle (theta) from the Horizontal. The Catch is that the place it was fired from is a hill sloped angle (phi) against the horizontal. The question is what are the angles are which you can get the highest Range (Range = not the Farthest Horizontal Distance, but the farthest distance along the slop).

Hint: If (phi) is 60, then (theta) is 15.

I then used the basic Kinematics equations and combined them. I solved for t under the horizontal displacement and plugged in what I got into my vertical displacement equation, but I keep going nowhere because I have a bunch of sines and cosines with theta and phi.

Hi CzarValvador! Welcome to PF! :smile:

(have a theta: θ and a phi: φ :smile:)

Show us your θ and φ equations … there's probably a simple trig trick for solving them (like the sin(A+B) rule). :wink:
 
Hey, thanks. Appreciate the help, but I was a little too late in figuring it out and turned it in. I had the physics TA solve it.

All he did was found where the equation of the height of the Slope, and an Equation for the height of the projectile and set them equal to each other.

Thanks for the help, expect more questions from me throughout the year :P
 
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