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A small cannonball weighing 5-kg is launched vertically in the air with an initial kinetic energy of 500 joules. How high will the cannonball travel above the launch point?

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Correct Answer: B. 10 m.
All of the kinetic energy will be transferred into potential energy at the maximum height of the cannonball. The trick is to not replace KE with (1/2)mv2, the kinetic energy is already given! Therefore:
KE -> PE -> KE = PE -> KE = mgh -> h = KE/mg = 500 J / (5 kg)(10 N/kg) = 10 m.
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