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Home Q & A Board Science-Related Homework Help Physics
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annadamere annadamere
wrote...
6 years ago
A 60 g ball is tied to the end of 50-cm-long string and swung in a vertical circle. The center of the circle, as shown in the figure below, 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?
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wrote...
#1
Educator
6 years ago
\(\frac{v^2}{r}=g\)

Solve for v.

\(v=\sqrt{g\cdot r}\)

= sqrt(9.8*0.5)

= 2.213 m/s

Let \(t\) is the time taken to hit the ground.

\(h=\frac{1}{2}g\cdot t^2\)

Solve for \(t\)

\(\sqrt{\frac{2h}{g}}=t\)

= sqrt(2*2/9.8)

= 0.639 s

So, \(x=v\cdot t\)

= 2.213 * 0.639

= 1.414 m
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