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OCW Scholar - Single Variable Calculus
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In the first problem set, Problem 1b-2 part c, the solution page does not show work. I've been working this problem over and over again and keep getting the same thing (-7b^2 / 32) and cannot figure out how they got (b^2 / 64) instead. Can anyone work out a solution?
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b^2/64 is correct. First take the derivative of s(t) to find the time of the maximum height. s'(t) = b-32t solving for 0 gives t=32/b. So the tennis ball is at maximum height at t = 32/b. To see how high it is at 32/b just plug it back to the expression s(t).
Duckbill - I see how you're getting your answer. You are calculating 16t^2 incorrectly. It's technically 16(t^2), which equals 16(b^2/1024), which equals b^2/64. You end up with b^2/32-b^2/64, which equals b^2/64.
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