Angular momentum of ball problem

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J-dizzal
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Homework Statement


In the figure, a 0.400 kg ball is shot directly upward at initial speed 40.4 m/s. What is its angular momentum about P, 6.65 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground?

Homework Equations


p=mv
l=r x p

The Attempt at a Solution


20150713_190337_zpsmmn73rsx.jpg

im stuck at part b, i used l=rxp but not sure if i did that right.
 
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Overall, your method looks good. However, your calculation of hmax is incorrect due to a sign error in the acceleration. A clue that something's wrong is that you got a longer time to fall halfway back down than the total time to reach maximum height. Also, your speed at half the max height is greater than the launch speed.

It might be easier to relate velocity directly to distance using vf2 = vi2 +2ay.
 
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You've gone wrong right near the start in calculating max height, because you've given g the same sign as the initial velocity, whereas in fact they are opposite. So your max height will be much lower.

There's a pretty big clue to this in that you calculate the speed when the ball's halfway back down as 49+ m/s, which is more than the initial speed, which is clearly impossible.

See if fixing up the errors helps you answer the question.
Edit: Jinx TSny
 
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thank you both. ill try again.

edit. that worked thanks!
 
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