# Solving for Theta

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Hi, So, given the formula for uneven ground here, I'm trying to solve for theta. Bear with me; I'm young yet. The result is not working properly (I've done something wrong somewhere): d = v*cos(θ)/g(v*sin(θ) + ((v*sin(θ))2+2*g*y0)0.5) => dg = v*cos(θ)*(v*sin(θ) + ((v*sin(θ))2+2*g*y0)0.5) => dg = v2*sin(θ)*cos(θ) + v*cos(θ)*((v*sin(θ))2+2*g*y0)0.5) => dg - v2*sin(θ)*cos(θ) = v*cos(θ)*((v*sin(θ))2+2*g*y0)0.5) => d2g2 - 2*d*g*v2*sin(θ)*cos(θ)+v4*sin2(θ)*cos2(θ) = v2cos2(θ)*(v2sin2(θ)+2*g*y0) => d2g2 - 2*d*g*v2*sin(θ)*cos(θ)+v4*sin2(θ)*cos2(θ) = v4*sin2(θ)*cos2(θ)+2*g*v2*y0*cos2(θ) => d2g2 - 2*d*g*v2*sin(θ)*cos(θ) = 2*g*v2*y0*cos2(θ) => g*(d2g-2*d*v2*sin(θ)*cos(θ)) = g*(2*v2*y0*cos2(θ)) => d2g - 2*d*v2*sin(θ)*cos(θ) = 2*v2*y0*cos2(θ) => d2g - 2*d*v2*sin(θ)*cos(θ) = 2*v2*y0*cos2(θ) => d2g = 2*v2*y0*cos2(θ) + 2*d*v2*sin(θ)*cos(θ) => d2g = 2*v2*(y0*cos2(θ)+d*sin(θ)*cos(θ)) => (d2g)/(2*v2) = y0*cos2(θ)+d*sin(θ)*cos(θ) => (d2g)/(2*v2) = y0*cos2(θ)+2*(1/2)*d*sin(θ)*cos(θ) => (d2g)/(2*v2) = y0*((1/2)*(1+cos(2θ)))+(1/2)*d*sin(2θ) => (d2g)/(v2) = y0*(1+cos(2θ))+d*sin(2θ) => (d2g)/(v2) = y0 + y0*cos(2θ) + d*sin(2θ) => ((d2g)/(v2)) - y0 = y0*cos(2θ) + d*sin(2θ) => Modifying this equivalence to handle equations of the y=a*sin(θ)+b*cos(θ) gives: => ((d2g)/(v2)) - y0 = (y02+d2)0.5sin(2θ+45°) => (d2g-2*v2y0)/(v2) = (y02+d2)0.5sin(2θ+45°) => (d2g-2*v2y0)/(v2(y02+d2)0.5) = sin(2θ+45°) => sin(2θ+45°) = (d2g-2*v2y0)/(v2(y02+d2)0.5) => 2θ+45° = arcsin( (d2g-2*v2y0)/(v2(y02+d2)0.5) ) => 2θ = arcsin( (d2g-2*v2y0)/(v2(y02+d2)0.5) )-45° => 2θ = arcsin( (d2g-2*v2y0)/(v2(y02+d2)0.5) )-π/4 => θ = 1/2arcsin( (d2g-2*v2y0)/(v2(y02+d2)0.5) )-π/8 Help? Thanks, -G

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As before, but a couple errors fixed, and in a much nicer format:
http://geometrian.com/Derivation.pdf
Thanks,
-G

##### Share on other sites
Again, as before, but a couple errors fixed:
http://geometrian.com/Derivation2.pdf
Thanks,
-G

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