The acceleration due to gravity on the Moon is 1/6th of what it is on Earth. If you can throw a baseball on Earth to reach a maximum height of 20 m, how high could you throw a base... The acceleration due to gravity on the Moon is 1/6th of what it is on Earth. If you can throw a baseball on Earth to reach a maximum height of 20 m, how high could you throw a baseball on the Moon? If you caught it when it came back down, how much time would elapse between your throw and catch? Read more

Steps to Solve

1. Understanding Gravity on Moon The acceleration due to gravity on the Moon is $\frac{1}{6}$ of that on Earth. Therefore, if the gravitational acceleration on Earth is approximately $g = 9.81 \text{ m/s}^2$, the gravitational acceleration on the Moon is: $$ g_{moon} = \frac{g}{6} = \frac{9.81}{6} \approx 1.635 \text{ m/s}^2 $$

2. Relation of Maximum Height and Gravity We can use the formula for maximum height in projectile motion: $$ h = \frac{v^2}{2g} $$ where $h$ is maximum height, $v$ is the initial velocity, and $g$ is the acceleration due to gravity.

3. Calculating Velocity from Earth's Height Given the maximum height reached on Earth is $h_{earth} = 20 \text{ m}$, we can rearrange the equation to find the initial velocity: $$ v_{earth} = \sqrt{2gh_{earth}} $$ Substituting the values: $$ v_{earth} = \sqrt{2 \cdot 9.81 \cdot 20} \approx \sqrt{392.4} \approx 19.8 \text{ m/s} $$

4. Calculating Maximum Height on the Moon Now, using this initial velocity on the Moon: $$ h_{moon} = \frac{v_{earth}^2}{2g_{moon}} $$ Substituting the Moon's gravity: $$ h_{moon} = \frac{19.8^2}{2 \cdot 1.635} \approx \frac{392.04}{3.27} \approx 120 \text{ m} $$

5. Calculating Total Time of Flight The total time of flight for an object in projectile motion where it goes up and then comes back down is given by: $$ T = \frac{2v}{g} $$ Using the initial velocity on the Moon, we find: $$ T = \frac{2 \cdot 19.8}{1.635} \approx \frac{39.6}{1.635} \approx 24.2 \text{ s} $$

6. Divide Time Measurement by 2 for Catch Time Since the total time includes both ascent and descent, the time when you catch the baseball is: $$ T_{catch} = \frac{T}{2} \approx \frac{24.2}{2} \approx 12.1 \text{ s} $$