Question

A person with a mass of 45 kg jumps off a diving platform that is 10.0 meters above the water. If they reach the surface of the water at a velocity of 12 m/s, what is the person's kinetic energy (KE)?
Hint: use the formula KE =12 mv2 (2 points)
Responses

5,400 J
5,400 J

32,400 J
32,400 J

72,900 J
72,900 J

3,240 J

Answer 1

To calculate the kinetic energy (KE) of the person just before they hit the water, you can use the formula:

\[
KE = \frac{1}{2} mv^2
\]

where:
- \( m \) is the mass (in kg),
- \( v \) is the velocity (in m/s).

Given:
- \( m = 45 \, \text{kg} \)
- \( v = 12 \, \text{m/s} \)

Now plug in the values:

\[
KE = \frac{1}{2} \times 45 \, \text{kg} \times (12 \, \text{m/s})^2
\]

Calculating \( (12 \, \text{m/s})^2 \):

\[
(12 \, \text{m/s})^2 = 144 \, \text{m}^2/\text{s}^2
\]

Now calculate the KE:

\[
KE = \frac{1}{2} \times 45 \times 144
\]
\[
KE = 22.5 \times 144
\]
\[
KE = 3240 \, \text{J}
\]

So, the person's kinetic energy just before hitting the water is **3,240 J**.

The correct response is:

**3,240 J**.