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 0.6m. The plane of the circle is 2.4 m above the ground. The string breaks and the ball lands 2.3 m away from the point on the ground directly beneath the ball's location when the string breaks. Use 10 N/kg for g and ignore air resistance. Show work please.

a) What was in meters the vertical displacement of the ball after it broke loose?

b) What was in meters the horizontal displacement of the ball after it broke loose?

c) For how long in seconds was the ball airborne after it broke loose?

d) What is in m/s the speed the ball had when it broke loose?

e) What was in m/s2 the centripetal acceleration of the ball during its circular motion.

Explanation / Answer

a)
Since the ball is in horizontal circle 2.4 m above the ground,
vertical displacement = 2.4 m

b)
Since ball lands 2.3 m from the point,
horizontal displacement = 2.3 m

c)
Use below equation in vertical direction:
d = vi*t+ 0.5*g*t^2
2.4 = 0+ 0.5*10*t^2
t=0.693 s

d)
Since there us no acceleration in horizontal direction,
Use below equation in horizontal direction:
d = vi*t
2.3 = vi * 0.693
vi = 3.32 m/s

Only 4 subparts at a time please

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