Question

During a practice dive, a 50-kg diver reached a maximum height of 4.6m above the water. She came to rest 0.42s after hitting the water.

Determine the average force that the water exerted on her while stopping her.

Express your answer to two significant figures and include the appropriate units. Enter positive value if the force is upward and negative value if the force is downward.

Answer 1

Answer:

F= + 1,131 kgF

Explanation:

performing an energy balance we have:

$mgh=\frac{1}{2}mv^2$ (1)

where:

m= mass of the diver = 50 kg

g = acceleration of gravity = 9.81 m/s2

h = maximum height = 4.6 m

V= speed reached before entering the water.

Simplifying ecuation (1):

$gh=\frac{1}{2}v^2$

$v=\sqrt{2gh}=\sqrt{2*9,81*4.6}= -9.50 m/s$ (downward)

To calculate the average force exerted on the diver, We will use te formula:

F=m*a

At the same time:

$a=\frac{V_f-V_0}{t}$

Where:

Vo= Speed before entering the water = -9.50 m/s

Vf = Speed after diver stops = 0 m/s

t= time to reach rest = 0.42 s

So;

$a=\frac{0-(-9.50)}{0.42}= + 22.62 m/s^2$ (upward)

F = m*a = 50kg*(+22.62m/s2) = +1131 kgF (upward)

Answer 2

Answer:

F = 1131 N

Explanation:

Given

Mass of the diver m = 50 kg

Maximum height reached by the diver H = 4.6 m

Time taken to come to come to rest after reaching water t = 0.42 s

Solution

Let us consider upward movements positive

The potential energy of the diver at the height H is converted into kinetic energy when he reaches the water surface

kinetic energy = potential energy

$\frac{1}{2} mv^2 =mgh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2 \times 9.8 \times 4.6} \\\\v = \pm 9.5 m/s^2$

Since the diver is moving downward

$v = -9.5 m/s^2$

The diver enters the water with this velocity and comes to rest inside the water

So change in velocity

$\Delta v = v_f - v\\\\\Delta v = 0 - ( - 9.5)\\\\\Delta v = +9.5 m/s$

Accleration

$a = \frac{\Delta v}{t} \\\\a = \frac{9.5}{0.42} \\\\a = 22.62 m/s^2$

Focre of the water

$F = ma\\\\F = 50 \times 22.60\\\\F = 1131 N$