Angular Momentum of Shot Question

[Ghostscythe]

Homework Statement

In the figure below, a 0.360 kg ball is shot directly upward at initial speed 37.5 m/s.

What is its angular momentum about P, 2.15 m horizontally from the launch point, when the ball is at the following heights?

• a) its maximum height
• b) halfway back to the ground

Homework Equations

\stackrel{\rightarrow}{l} = m(\stackrel{\rightarrow}{r} \times \stackrel{\rightarrow}{v}), which given this problem simplifies to l = r_{\bot}mv
v^2 = v^{2}_{0} + 2a(x - x_{0})

The Attempt at a Solution

For a), v was obviously 0 at max height, so the angular momentum was 0 as well.

For b):
First, I needed to find the halfway position in regards to y, to help find the velocity there. Taking:
* v_{initial} = the time just after being launched from the ground
* v_{final} = the apex of its climb

I plugged in the numbers: 0^2 = 37.5^2 + 2(-9.8)(y - 0) and got: y = 71.747

Now using the same equation, except with v_{final} being halfway up and y obviously half of what it was:
v^2 = 0^2 + 2(9.8)(35.874 - 0)

I got v^2 = 703.125 so v = -26.517 m/s.

Edit: Plug in the numbers, and (-26.517 m/s)(2.15 m)(0.36 kg) = 20.52 N*m*s, which is right.

Not sure what I did the first time, but thanks, I guess having to do the post format for this forum helps, haha!

 

[Chi Meson]

Glad we could help!