- #1

ScreamingIntoTheVoid

## Homework Statement

In the figure, a 0.400 kg ball is shot directly upward at initial speed 58.3 m/s. What is its angular momentum about

*P*, 1.42 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

v^2=(v0)^2 -2gx

L=m(r x v)

J= F x r

## The Attempt at a Solution

I got a,c, and d right (I'm including them just in case some other confused soul passes by and needs a hint), but I couldn't get b.

a) 0 because velocity is 0

b)

Max hight: v^2=(v0)^2 -2gx --> 0 = (58.3^2) -2(9.8)x --> 173.4127551 m -->divide by 2 to get half height = 86.70637755 m

again, v^2=(v0)^2 -2gx --> v^2=0 - 2(9.8)(86.70637755) --> -39.50588305 m/s [assuming the math is right]

L=mrv --> (0.4kg)(-39.505m/s)(1.42m) --> -22.43934157 kg*m^2/s

**[This is wrong even if I just do magnitude]**

c and d) f x r=J --> (fx)(ry)-(fy)(rx)= J ~> (got right so the answer doesn't matter)

Can anyone tell me what went wrong in problem b? Thanks in advance to anyone who responds