Question
A boulder with a mass of 1.35 kg is dropped off of a 207 m cliff. What will the speed of
the boulder be a split second before hitting the ground?
the boulder be a split second before hitting the ground?
Answers
potential energy becomes kinetic energy
m g h = 1/2 m v^2
v = √(2 g h) = √(9.8 m/s^2 * 270 m) = ? m/s
m g h = 1/2 m v^2
v = √(2 g h) = √(9.8 m/s^2 * 270 m) = ? m/s
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