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The ball has 7.35 joules of potential energy at position B. At position A, all of the energy changes to kinetic energy. The velocity of the ball at position A is meters/second. Assume there’s no air resistance. Use g = 9.8 m/s2 , PE = m × g × h, and . mass is 1.5 kilograms and g= 9.8 and the hight of position b is 0.5 meters.

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skyluke89
I assume that the ball is stationary (v=0) at point B, so its total energy is just potential energy, and it is equal to 7.35 J. 
At point A, all this energy has converted into kinetic energy, which is:
[tex]K= \frac{1}{2}mv^2 [/tex]
And since K=7.35 J, we can find the velocity, v:
[tex]v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 7.35 J}{1.5 kg} }=3.1 m/s [/tex]

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