<p>Hi all. I was just wondering if anyone might be able to help me out with the question located at the link below...not just the answers (I already have them), but how you arrived there.</p>
<p>Thanks</p>
<p>Hi all. I was just wondering if anyone might be able to help me out with the question located at the link below...not just the answers (I already have them), but how you arrived there.</p>
<p>Thanks</p>
<p>A) I think it would just be 1/2mv^2+mgh=2.78 J. It shouldn't matter where the ball is located.<br>
B) If the energy remains constant than 1/2mv^2+mgh=2.78 J. Plug in the values at the bottom of the circle, h=.2 instead of 1. V=7.188 m/s
C)
i. T+mg=(mv^2)/r, T=5.48 N
ii. T-mg=(mv^2)/r, T=7.438 N
D) Its flight will be simply a parabola so you can just use kinematics. y=vit+1/2at^2 where vi is zero, t=.202 s. Then you can use the same equation but in the x direction where a=0 so x=vit, x=1.45 m.
I'm not sure if these are right. Didn't check the answers and I'm not a physics whiz or anything.</p>
<p>Yeah, I got all those answers, too. I had all my work on a peice of paper and I was going to scan it but then I forgot about it. yay.</p>