- #36
Delphi51
Homework Helper
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- 11
It just isn't that simple. In the rotated system you have accelerated motion in both directions. Consider the case when alpha is 80 degrees. Your answer says shoot at pi/4 + alpha = 45+80 = 125 degrees, which is actually away from the hill and would never hit it.
RoyalCat's idea of using x instead of r seems to reduce the algebra considerably. I worked it through to the end and got the same answer I got last night with the r. Unfortunately there seems to be an error in my work. I get tan(2θ) = - 1/tan(alpha) with a condition alpha not equal to zero. So when alpha is 10 degrees, it says shoot at -40 degrees.
Edit: I used a spreadsheet to take a good look at my solution
x = 2*v^2/g(sinθcosθ-tanα*cosθcosθ)
before taking the derivative to get the max, and it works. With alpha 10 degrees, shooting at 48 degrees gives the max range. So I just have an error in the derivative or last bit of algebra.
Recommend you use y = tan(alpha)*x to eliminate y in the original equation y = v*sinθ*t -.5*g*t^2 and use x = v*cosθ*t to eliminate t. Solve for x. About 3 lines on you'll see my x = expression above.
RoyalCat's idea of using x instead of r seems to reduce the algebra considerably. I worked it through to the end and got the same answer I got last night with the r. Unfortunately there seems to be an error in my work. I get tan(2θ) = - 1/tan(alpha) with a condition alpha not equal to zero. So when alpha is 10 degrees, it says shoot at -40 degrees.
Edit: I used a spreadsheet to take a good look at my solution
x = 2*v^2/g(sinθcosθ-tanα*cosθcosθ)
before taking the derivative to get the max, and it works. With alpha 10 degrees, shooting at 48 degrees gives the max range. So I just have an error in the derivative or last bit of algebra.
Recommend you use y = tan(alpha)*x to eliminate y in the original equation y = v*sinθ*t -.5*g*t^2 and use x = v*cosθ*t to eliminate t. Solve for x. About 3 lines on you'll see my x = expression above.
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