A 73.0 kg diver steps off a diving board and drops straight down into the water.

Question

The water provides an upward average net force 1390 N. If the diver comes to rest 5.9 m below the water's surface, what is the total distance between the diving board and the diver's stopping point underwater?

Answer 1

As the average force in stopping the diver’s descent in the water is 1390 N and the mass is 73.0 kg, the deceleration (negative acceleration) is a = F / m = 1390 N / 73.0 kg = 19.0 m/s^2. Let V_s be the velocity with which the diver hits the water’s surface. The V_s is given by 0 = V_s² – 2aD where D is the depth to which the diver sank (= 5.9 m) and a is 19.0 m/s². Solving for V_s yields 14.97 m/s.

Since the diver started from zero velocity, the equation V_s² = 0² + 2gH where H is the height from the water to the board and g the acceleration of gravity (9.8 m/s²).  Solving for H gives 11.4 m. The total distance is the sum of 5.9 m + 11.4 m = 17.3 m.

Answer 2

The work done by the water is F•d = 1390•5.9 = 8201 Nm

All the kinetic energy that the diver had was converted to work so the kinetic energy (at the point the diver enters the water) = 8201 joules.

All the potential energy that the diver had when they were on the diving board was converted into kinetic energy so the potential energy (E_P) = 8201 joules

E_P = mgh

8201 = (73.0)(9.81)h

h = 11.5 meters

Since the question wants to know the total distance between the diving board and the diver's stopping point underwater, then we must add the 11.5 (out of the water) with the 5.9 m (under the water) to get 17.4 meters