A 2 kilogram rock is a the edge of a cliff 20 meters above a lake. The rock becomes loose and falls toward the water below. What is the potential energy of the rock at the top? What is the kinetic energy at the top?

What is the potential enery of the rock half way down?

What is the kinetic energy of the rock half way down if the velocity at that point is 14 m/s?

The answer would be d. hope this : ))

En espanol amigo por favor

Answer(550) lb (1) foot each second

A) the moment of inertia is given by i = mr^2 here m = 0.25 and r = 0.5. i = 0.25 * (0.5)^2 = 0.0625. b) i = 9.74 * 10^(37) and mass = 5.98 * 10^(24) so r^2 = i/m = 9.74 * 10^(37)/5.98 * 10^(24) = 1.628 * 10^(37 - 24) = 1.628 * 10^(13) r = √ 1.628 * 10^(13) = 4034848.200 c) since angular momentum is the product of moment of inertia and angular velocity, a = i * v; but i = mr^2 = 2 * 0.6^2 = 0.72 so a = 0.72 * 1.1 = 0.792. if she pulls her hand in 0.15 m. the new radius = 0.6 - 0.15 = 0.45. v = angular momentum/ radius = 0.792/ 0.45 = 1.76