Originally posted by SyedAshrafulla
It's definitely way dependent on the velocity magnitude v, because the goal posts are 10 feet high (so the time of flight is more confusing). Normally in your 45-degree estimate, it's b/c v t sin Θ - 1/2 a t^2 = 0, so t = 2v/a sin Θ and thus the distance covered is vt cos Θ = v^2/a sin 2Θ which maximizes over Θ when 2Θ = π/2, or Θ = π/4 (45 degrees).
The problem is that now you constrain v t sin Θ - 1/2 a t^2 = 10, which means the ball meets the plane of the uprights at a much more complicated expression: t = (v sin Θ + sqrt(v^2 sin^2 Θ - 20g))/a and thus we want to maximize, over Θ, the function d(Θ = v/a cos Θ (v sin Θ + sqrt(v^2 sin^2 Θ - 20g)) = v/a (v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). Notice that if we were dealing with no crossbar, the 20g would go away and we'd get the above solution. We can ignore the v/a too.
So the summary is that the optimal Θ is one that maximizes v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). I tried to tkae the derivative, etc. and it turns out to be a cubic equation in the variable sin^2 Θ, which means it's solvable and probably only has one solution in [0, π/2].
wat
It's definitely way dependent on the velocity magnitude v, because the goal posts are 10 feet high (so the time of flight is more confusing). Normally in your 45-degree estimate, it's b/c v t sin Θ - 1/2 a t^2 = 0, so t = 2v/a sin Θ and thus the distance covered is vt cos Θ = v^2/a sin 2Θ which maximizes over Θ when 2Θ = π/2, or Θ = π/4 (45 degrees).
The problem is that now you constrain v t sin Θ - 1/2 a t^2 = 10, which means the ball meets the plane of the uprights at a much more complicated expression: t = (v sin Θ + sqrt(v^2 sin^2 Θ - 20g))/a and thus we want to maximize, over Θ, the function d(Θ = v/a cos Θ (v sin Θ + sqrt(v^2 sin^2 Θ - 20g)) = v/a (v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). Notice that if we were dealing with no crossbar, the 20g would go away and we'd get the above solution. We can ignore the v/a too.
So the summary is that the optimal Θ is one that maximizes v/2 sin 2Θ + cos Θ sqrt(v^2 sin^2 Θ - 20g)). I tried to tkae the derivative, etc. and it turns out to be a cubic equation in the variable sin^2 Θ, which means it's solvable and probably only has one solution in [0, π/2].
wat