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## Homework Statement

A 60 g ball is tied to the end of a 40-cm-long string and swung in a vertical circle. The center of the circle 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, how far to the right does the ball hit the ground?

## Homework Equations

vy^2 = voy^2 + 2a(y-yo)

vy = voy + at

ac=vx^2/r

(x-xo) = voxt + (1/2)axt^2

## The Attempt at a Solution

I used vy^2 = voy^2 + 2a(y-yo) to find the final velocity in the y direction :

vy^2 = (0)^2 + 2(-9.8)(-0.19)

vy = 1.93 m/s

Then, I used that final velocity and found the time it took to hit the ground :

vy = voy + at

(1.93) = (0) + (9.8)t

t = 0.1969 s

After that, I used the ac=vx^2/r equation to find the velocity in the x direction :

(9.8) = (vx^2)/(0.4)

vx = 1.98 m/s

Then, I used my velocity in the x direction, my time, and my zero acceleration in the x direction to find my delta x :

(x-xo) = voxt + (1/2)axt^2

(x-xo) = (1.98)(0.1969) + (1/2)(0)(0.1969)^2

(x-xo) = 0.3899 m OR 38.99 cm

The online program told me that it was wrong and all of my friends at my department don't know where I went wrong, can someone please help me?

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