A 67.0 kg diver steps off a diving board and drops straight down into the water. The water provides an upward average net force 1480 N. If the diver comes to rest 4.7 m below the water's surface, what is the total distance between the diving board and the diver's stopping point underwater?

Answer 1

the upward acceleration from the water is a=F/m = 1480/67 = 22.09 m/s^2

The diver's speed on entering the water is

v = √(2as) = √(2*22.09*4.7) = 14.4 m/s

In the air, the diver's speed is

v = 4.9t^2, so he fell for

t = √(14.4/4.9) = 1.71 seconds

The diver's height upon diving is

h = Ho - 4.9t^2
h=0 at t=1.71, so

Ho = 4.9*1.71^2 = 14.3

The total distance is thus 14.3+4.7 = 19.0 meters

Answer 2

the diver's air speed is v = 9.8t

make that correction and redo the board height.