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Part A l. A ball is thrown vertically, starting at height 2 = 0 with speed 1),, = 20mfs. It should be clear that the height as a
function of time will be 2(t) = vet — 1/2915?
i) Calculate the time T it takes to return to z = 0, and calculate the coefﬁcients of the Fourier series, assuming that the motion is periodic (as would happen if there were an ideal trampoline at z = 0). Shift the interval to t =
[—T/ 2,172], and use the results from homework 6 for the Fourier series of a parabola (but subtract 0.25 from that answer, ﬂip the sign of everything, and then rescale to get the right height). Recall that the solution in that case was
an 2 1/(mr)2(—1]&quot;. ii) Write an expression for the action S = f L(z, :3), using this Fourier series, integrating from the time the ball is
launched to the time it returns to the ground. iii) Truncate the Fourier series at n = 3, and instead of using the correct n = 3 term for the expansion, treat it as a
variable, and plot the action as a function of a3.

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