A ball on the end of a string is whirled around in a horizontal circle of radius

Question

A ball on the end of a string is whirled around in a horizontal circle of radius of 0.317m. The plane of the circle is 1.82m above the ground. The string breaks and the ball lands 2.94 m away from the point on the ground directly beneath the ball's location when the string breaks. The acceleration of gravity is 9.8 m/s^2. Find the centripetal acceleration of the ball during its circular motion.

I know that ultimately I will use Ac = v^2 / r. But I'm not seeing how to find velocity so I can use the centripetal acceleration equation. Any help would be appreciated.

Explanation / Answer

1.82 m/.5(9.8m/s2) = t2 t = .6094s Traveled 2.94m Velocity in the x dimension doesn't change

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.