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anonymous
 5 years ago
In the Clip 5: Ballistic Pendulum (12 min) of Momentum of the MIT classes, he calculates “h” using a approximation of h=x2/2L. I don’t understand why he does that, he could simply use Pythagoras to find h0 and then h = 1mh0.
Please, if someone could help me I would appreciate it.
anonymous
 5 years ago
In the Clip 5: Ballistic Pendulum (12 min) of Momentum of the MIT classes, he calculates “h” using a approximation of h=x2/2L. I don’t understand why he does that, he could simply use Pythagoras to find h0 and then h = 1mh0. Please, if someone could help me I would appreciate it.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0He starts off using the Pythagorean theorem, but points out that h is very difficult to measure when the angle is only 2 degrees. He goes on to use x^2/2L because one can measure x with much less relative error. The only reason is to keep the error in the answer small (which is one was Physics differs from mathematics  understanding error is important!)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I understand why he messures X insted of h, is very clever. What confuses me is why he uses x^2/2L instead of use Pythagorean with: h=100ROOT(100^23,5^2)=6,127mm, which is a exact result (no an approximation) and uses a more familiar formula. (The other method gives 6,125mm). Thanks
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