Question

A cliff diver with a mass of 175 lbm dives from a 20 foot cliff into the sea. The initial velocity of the diver as he/she dives off the cliff is 5 ft/s. Assume gL = 32.2 ft/s². Determine the following: a) The potential energy of the diver when the diver reaches a velocity of 20 ft/s. b) The height of the diver above the surface of the water when the diver reaches a velocity of 20 ft/s. c) The velocity of the diver at the first contact with the water.

Answer 1

a) The potential energy of the diver when the diver reaches a velocity of 20 ft/s is 112,700 ft·lb. b) The height of the diver above the surface of the water when the diver reaches a velocity of 20 ft/s is approximately 19.2 feet. c) The velocity of the diver at the first contact with the water is approximately 18.9 ft/s.

Step-by-step explanation:

a) To determine the potential energy of the diver when the diver reaches a velocity of 20 ft/s, we need to use the equation for potential energy: PE = mgh. Given that mass of the diver is 175 lbm and the acceleration due to gravity is 32.2 ft/s², and the height is 20 feet, we can calculate the potential energy using the equation: PE = (175 lbm)(32.2 ft/s²)(20 ft) = 112,700 ft·lb.

b) The height of the diver above the surface of the water can be found using the equation for potential energy: PE = mgh. Rearranging the equation, we have: h = PE / (mg). Substituting the given values, we get: h = (112,700 ft·lb) / (175 lbm)(32.2 ft/s²) ≈ 19.2 feet.

c) The speed of the diver at the first contact with the water can be found using the equation for kinetic energy: KE = 1/2 mv². Rearranging the equation, we have: v = √(2KE / m). Substituting the given values, we get: v = √(2(112,700 ft·lb) / (175 lbm)) ≈ 18.9 ft/s.

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