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## Homework Statement

A 60 g ball is tied to the end of a 50-cm-long string and swung in a vertical circle. The center of the circle, as shown in Figure, is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack.

If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?

http://session.masteringphysics.com/problemAsset/1000600/5/knight_Figure_07_62.jpg

## Homework Equations

Critical velocity = [tex]V{c}[/tex] = [tex]\sqrt{r*g}[/tex]

[tex]\Delta[/tex]d=V{i}*t + (1/2)(a)(t^2)

## The Attempt at a Solution

I attempted this problem as a projectile motion problem, and divided the calculations into two parts: Horizontal and Verticle

**Verticle Part**

So I used this equation: [tex]\Delta[/tex]d=V{i}*t + (1/2)(a)(t^2) and I know the verticle initial is 0 so I can eliminate V{i}*t of the equation and get: [tex]\Delta[/tex]d=(1/2)(a)(t^2)

2 = (1/2)*(9.8)*(t^2)

2 = 4.9*(t^2)

t^2 = 2 / 49

t = 0.64 seconds

**Horizontal Part**

The horizontal parts only has 3 variables: V, d, and t (which I know from the verticle part). and V = V{c} = [tex]\sqrt{r*g}[/tex] = 0.59 m/s.

and d = v*t = 0.59*0.64 = 0.3776 m

This doesnt look right. I was wondering if someone can point me in the right direction. Any help is greatly appreciated. Thanks.

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