Question
**THE EXAMPLE:
##THE QUESTION:
A 60g ball is tied to the end of a 60-cm-long string and swung in a vertical circle. The center of the circle, as shown in the figure, (Figure 1) 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. Figure 1 of 1 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? Express your answer to two significant figures and include the appropriate units. Delta x = 1.6 m Submit My Answers Give Up Answer Requested A 60g ball is tied to the end of a 70-cm-long string and swung in a vertical circle. The center of the circle, as shown in the figure, (Figure 1) 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. Figure 1 of 1 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? Express your answer to two significant figures and include the appropriate units. Delta x = Submit My Answers Give Up
Explanation / Answer
velocity of the ball at instant it is left can be calculated as
mg=mv2/r
implies v=(rg) = (60*9.8) = 24.24m/s
now with intial velocity 24.24m/s the horizontal distance ball will travel
=v*time where time(t) is time taken by ball to reach ground
t=(2*210m/9.8){ bcoz s=1/2gt2} = 6.54secs
implies horizontal distance = 24.24*6.54m= 158.68m
x=158.68m
