Question

what is the kinetic energy of a 740 kg sky diver falling at a terminal velocity of 52 m/s?

Answer 1

What is the kinetic energy of a 740 kg sky diver falling at a terminal velocity of 52 m/s?

Solution.

The formula for kinetic energy in general form is:

$$E = \frac{m v^{2}}{2},$$

where $m$ is the mass of an object, $v$ is the velocity of an object, $E$ is the kinetic energy of an object.

So we have:

$$E = \frac{m v^{2}}{2} = \frac{740 \cdot 52^{2}}{2} = 1000,48 \text{ kJ}.$$

Answer: 1000,48 kJ.