Asked by patti
If a coyote is standing at the top of a 125 m cliff getting ready to drop a 230 kg boulder on a road runner. What is the potential energy of the boulder just as he is ready to drop it? What is the kinetic energy of the boulder at the instant it hits the ground? How fast is the boulder going at the instant it hits the ground? I need a step by step formula for the questions and the answers.
Answers
Answered by
Henry
Ep = mg*h = 230*9.8*125 = 281,750 Joules
Ek = Ep = 281,750 Joules.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*125 = 2450
V = 49.5 m/s.
Ek = Ep = 281,750 Joules.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*125 = 2450
V = 49.5 m/s.
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