- #1

amcavoy

- 665

- 0

_{o}with an initial velocity of 0. I want to find the velocity of the marble at any height along the way. What I've done is below:

[tex]mgh=\frac{1}{2}mv^{2}\implies v=\sqrt{2gh}[/tex]

Since g is downward, I worked it out that the force in the direction of motion would be [itex]g\sin{\theta}[/itex]. Thus the velocity at any given height would be [itex]\sqrt{2gh\sin{\theta}}[/itex]. However, this doesn't account for the fact that the ball begins at rest. Is this incorrect or am I on the right track?

Thanks for your help.