Question

A 65 kg person dives into the water from the 10 m platform. (h as shown in diagram.) He comes to a stop 2.9 m below the surface. (d as shown in diagram.) Note: we are ignoring air friction here, but not ignoring water friction.
(a) What is his speed as he enters the water? (do this as if he falls off the edge of the board like a stone.)
(b) What is the magnitude of his acceleration while in the water?
(d) What is the total amount of force exerted on the swimmer in the water?

Answer 1

a ) v = v o + g t
h = v o * t + g t² /2
10 m = 9.81 m/s² * t² / 2
9.81 t² = 20
t² = 2.038
t = 1.4278 s
v = 0 + 9.81 * 1.4278
v = 14 m/s
b )  v o = 14 m/s
v = v o² + 2 a d
0 = 14² + 2 a * 2.9
196 = - 5.9 a
a = 196 : ( - 5.9 )
a = - 33.79 m/s² ( the magnitude of his acceleration while in water )
c ) The total amount of force exerted on the swimmer in water:
F = m * a = 65 kg * 33.76 m/s²
F = 2,196.55 N