Recent content by jmwachtel

No it would be between 30 and 100 I would assume. Would it just be in the middle of that? I'm not following...

I was just using the 1/12 formula? Does that calculate it at 50cm?

Homework Statement A particle of mass 0.500 kg is attached to the 100 cm mark of a meter stick of mass 0.175 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 5.00 rad/s. (a) Calculate the angular momentum of the system when the stick is pivoted...

Ok, then what do I need to solve the problem? Does the angle matter?

I'm still lost?

Well I am looking for the potential energy of the 1st ball on part b which has the 42 degree angle. All of the energy is still potential right?

so I should be looking for mg*Vyxi?

Is this correct?

Ok, so I would use mg (Vsin(42))?

I see on the 2nd ball that is fired the potential energy is 1.10e7 which is just mgy. How is the 1st ball calculated? I guess that's where I am having the most problem with that angle. Is it still just mgy?

How do you find velocity in the vertical direction?

Homework Statement A 20.0 kg cannonball is fired from a cannon with a muzzle speed of 1050 m/s at an angle of 42.0° with the horizontal. A second ball is fired at an angle of 90.0°. (a) Use the isolated system model to find the maximum height reached by each ball. (b) What is the...

Anyone understand this?

Ok here is my teachers reply: Your integration is still incorrect. See attachment for explanation. You MUST have correctly corresponding limits on the integrals. At time=0, y=yo; at time=t, y=y. I don't understand. The attached says: v = dy/dt dy = vdt \intdy (from yintial to...

[\frac{mgt}{b} + \frac{mg}{b} (\frac{m}{b} e^{\frac{-bt}{m}})] - [\frac{m^{2}g}{b^{2}}] This is integration I have come up with for the equation listed above that is not working! IS my integration wrong? This from 0 to t.