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November 26, 2014

November 26, 2014

Posted by **jane** on Saturday, June 13, 2009 at 7:28pm.

I know that because max height is v=0, there is 0 angular momentum, but i can't figure out the rest

- physics -
**Damon**, Saturday, June 13, 2009 at 7:57pmDo the second question first. (It is worded incorrectly I think because there is in fact no torque on the ball since the force goes right through its center. There is however a torque about that point 2 meters from the launch point) The torque is the same at the top and halfway down because the perpendicular distance from the force vector to the launch point is always 2 meters. The magnitude of the torque = m g (2)

Well, we better figure out how fast it is going when it is halfway down. You already answered for at the top, zero because velocity is zero.

a = - 9.8

v = 40 - 9.8 t

h = 0 + 40 t - 4.9 t^2

when h = 20

20 = 40 t - 4.9 t^2

4.9 t^2 - 40 t + 20 = 0

t = (1/9.8)[ 40 +/- sqrt(1600 -392)]

t = (1/9.8) [ 40 +/- 34.8 ]

t = 7.63 on the way down (the plus sign)

v = 40 - 9.8(7.63) = -34.8 m/s

angular momentum = m V x R

= m (-34.8)(horizontal distance which is 2)

= -69.6 * mass

- physics -
**jane**, Saturday, June 13, 2009 at 8:28pmTHANK YOU SO MUCH!

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