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A 91 kg kangaroo exerts enough force to acquire 2.7 KJ of kinetic energy in jumping straight upward. a) Apply the law of conservation of energy to determine how high this agile marsupial jumps. b) What is the magnitude of the kangaroo's maximum velocity?
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Kinetic energy, KE, at the beginning of the jump is all converted into gravitational potential energy, PE. Now \( KE = \frac{1}{2}mv^2 \), where mass of the kangaroo is m and the speed is v. What is the expression for gravitational PE as a function of m, mass g, the gravitational acceleration h, the height ? When you know what that is, set PE = KE, as given, i.e., KE = 2700 J and solve for height h.
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