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) 14 meters/second

Answer 2

Hmm, I don't think there is sufficient information for (b). Is there another part of the problem you've left out (e.g. from an earlier problem that was related)?

Answer 3

Oh wait, never mind... I got it

Answer 4

(b) is 33.79 m / s^2

Answer 5

Are you familiar with the constant acceleration equations of motion?
\[\left(\begin{matrix}v = at+v_o & (1) \\ v^2 = v_o^2 +2a(x-x_o) & (2)\\ x = {1 \over 2} at^2 +v_ot +x_o & (3)\end{matrix}\right)\]
Let's use equation 2 here to find the velocity as the diver reaches the water. \[v^2 = (0)+2(9.81)(10-0)\]We can use equation 2 to find the acceleration in the water\[(0) = v^2 + 2a_w(2.9)\]where v is the velocity as he enters the water. Force is mass times acceleration. \[F_w = m_d a_w\]

Answer 6

(c) is 2196.35 newtons