To solve this problem, we can use the concept of energy conservation.

When the person jumps, their kinetic energy at the start is converted into gravitational potential energy at the maximum height.

The formula for kinetic energy is: $KE = \frac{1}{2} m v^2$ where:

• $m$ is the mass of the person (60 kg),
• $v$ is the speed they leave the ground with (3 m/s).

The formula for gravitational potential energy is: $PE = mgh$ where:

• $g$ is the gravitational field strength (9.8 N/kg),
• $h$ is the maximum height.

At the highest point, all the kinetic energy has been converted to potential energy: $\frac{1}{2} m v^2 = mgh$

We can cancel out $m$ from both sides: $\frac{1}{2} v^2 = gh$

Now, solve for $h$: $h = \frac{v^2}{2g}$

Substitute the values: $h = \frac{(3)^2}{2 \times 9.8} = \frac{9}{19.6} \approx 0.46 \text{ meters}$

Thus, the maximum height the person can reach is approximately 0.46 meters (to 2 significant figures).

Would you like more details or have any questions?

Here are 5 related questions you might want to explore:

1. How would the height change if the person’s speed was doubled?
2. How does air resistance affect the maximum height in real-world scenarios?
3. What would happen to the height if the person’s mass increased?
4. Can we calculate the time it takes for the person to reach the maximum height?
5. How is energy conservation applied in other physical scenarios?

Tip: When solving energy problems, remember that energy is always conserved in closed systems, making energy equations powerful tools for calculations!