# Angular momentum in projectile motion

1. Apr 19, 2010

### mybrohshi5

1. The problem statement, all variables and given/known data

A rock of mass 55.0 g is thrown with initial horizontal velocity v_x = + 32.00 m/s off a building from a height of 29.00 m. Take UP to be the +y direction. You will need to calculate the angular momentum of the rock about the line along the edge of the roof as a function of time.

Find the angular momentum of the rock about the line along the edge of the roof at 0.28 s.

2. Relevant equations

L = r*m*v

3. The attempt at a solution

to find the distance traveled in .28 seconds

d = 32m/s(.28) + .5(9.8m/s^2)(.28^2)
d = 9.344 m

then L = r*m*v

L = 9.344(.055kg)(32m/s)

I know this is wrong but i am not sure how else to go about this problem :(

thank you for any help

2. Apr 20, 2010

### ehild

The angular momentum is a vector product:

$$\vec L = m \vec r \times \vec v$$

For motions in the xy plane, L has only z component and it is

$$L = m ( x v_y - y v_x )$$.

Determine the x, y coordinates of the projectile as function of time with respect to the edge as origin, and the x and y components of its velocity too, and substitute in the above expression for L.

ehild