Question

A swimmer of mass 58 kg steps off a diving board 15 m above the water. Six seconds after entering the water, her downward motion is stopped. What average upward force did the water exert on her?

Answer 1

165.88 N

Explanation:

We have that when it enters the water, the upward force exerted on the swimmer slows it down and stops it. Its speed when it enters the water can be determined using an appropriate equation of motion. The average force experienced by the swimmer can be determined by applying Newton's second law to the swimmer's motion.

The swimmer starts with his initial velocity equal to zero. We calculate its speed v when it reaches the surface of the water, using the following expression:

`v^2=u^2-2\cdot g\cdot h`

Given:

u = 0

g = 9.8 m/s^2

h = -15 m

Replacing:

`\begin{gathered} v^2=0^2-2\cdot9.8\cdot-15 \\ v^2=294 \\ v=\sqrt[]{294} \\ v=\pm17.15\text{ m/s} \end{gathered}`

The swimmer's speed when he hits the water is directed downward, so the negative sign is chosen for the speed v.

v = -17.15 m/s

The swimmer enters the water and is slowed down by the drag exerted by the water on her and stops at a time t. Its average acceleration during the descent time is given by the expression:

`a_(av)=(u-v)/(t)`

Given:

u = 0 m/s

v = -17.15 m/s

t = 6 sec

Replacing:

`\begin{gathered} a_(av)=(0-(-17.15))/(6) \\ a_(av)=2.86m/s^2 \end{gathered}`

Now, we apply Newton's law to the swimmer's motion:

`\begin{gathered} F_(av)=m\cdot a_(av) \\ \text{ Given:} \\ m=58\text{ kg} \\ a_(av)=2.86m/s^2 \\ \text{ Replacing} \\ F_(av)=58\cdot2.86 \\ F_(av)=165.88\text{ N} \end{gathered}`