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September 20, 2014

September 20, 2014

Posted by **Sandhya** on Monday, October 27, 2008 at 11:30pm.

I used energy consideration & got the following answer. But the answer is incorrect.Please point out where i have gone wrong

Under the water you have the net force on the ball; F= 40 N. That force acts over

a distance D (1.5 m in this case) and thus does work on the ball giving it some

KE as it leaves the water: KE = FD. = 40*1.5 = 60 J

To find out how high the ball rises in the air, set that initial KE to the PE at the top

of the motion: KE = FD = 60 J = mgH. Now solve for H. where m = 1 kg

H = 60/1*9.8

= 6.12 m

- physics -
**drwls**, Tuesday, October 28, 2008 at 7:39amPerhaps it has something to do with the radius of the ball. When the center of the ball has risen 1.50 - 0.44 m, it is no longer fully submerged. They want you to assume that work stops then. It is not clear where to measure the potential energy change from: the surface or one radius below.

The answer you get depends upon the approximations made

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