Bob A. answered • 08/06/14

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For part A you have ignored the potential energy at the starting position of 1.5 m above ground.

U

_{i}= mgh = (0.005kg) (9.81 m/s^{2}) ((1.5 m) =For part B finding the max position is needed as you do.

But not that way. You have U

^{2}- v^{2 }How can you subtract velocity from energy?Since at the top v

_{f}= o we can take advantage of that and the fact that the up and down problems at symmetrical/mirror images with the falling velocity = initial velocity at the same height.Δx = v

_{f}^{2}/ 2a So you can see you were doing the correct maths with numbers but but not from the right formula.Now for the second part of B

U = mgh that's right, but h is the 20.39 m + 1.5 meters ot less than less than

So for a) K = the 1 J , and U = mgh where h is 1.5 m - then total them.

and for b) K = 0 (zero v at top) and U is mgh with h = (20.39 m + 1.5 m)

Haizal O.

08/06/14