Read please example 2 of http://cnx.org/content/m14060/latest/ .
P1
P2
On picture 2 there are those two tension labeled by T which are equal. Why equal?
And shouldn't they exert just torque? I know we are considering the pulley to be massless, so no torques. So is it correct to...
This might help: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra14
\frac{h}{R} = \frac{\tan \theta}{4} = 1/2
Then, you may find theta.
Note: You should not memorize this formula, but understand how to deduce it.
You have data to calculate the center of mass before the people move. Since, there are no external forces acting on the c.m., will he change?
So,
x_{cm}_{i} = x_{cm}_{f}
I think I will land somewhere backwards, there's no force acting on me so the train during that time is traveling faster than me. But I could be wrong, that is why I asked. :wink:
Yes, you did it correctly. Remember that you could calculate the displacement averaging out the velocity only because the acceleration is constant.
The other formula would be: d = (1/2)at2. Check that the solution is the same.
Try to show that v(average)*t = (1/2)at2
Visual1Up, I think you know how to derive the formula that gives range in function of \theta , v_o, which is:
R = (v_0^2 \sin 2\theta)/g
Then, it's easy. :smile: