The acceleration due to gravity on the Moon is 1/6 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 baseba... The acceleration due to gravity on the Moon is 1/6 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?
Understand the Problem
The question is asking about the effects of gravitational differences between the Earth and the Moon on the height that can be achieved when throwing a baseball. It requires calculations based on the maximum height reached on Earth and the comparison to the same action on the Moon, considering the differing gravitational accelerations. Additionally, it asks to determine the time elapsed from the throw until the baseball is caught again.
Answer
The maximum height on the Moon is approximately $120.0 \, \text{m}$, and the time elapsed is approximately $24.23 \, \text{s}$.
Answer for screen readers
On the Moon, you could throw the baseball to a maximum height of approximately $120.0 , \text{m}$, and the time elapsed from throw to catch would be approximately $24.23 , \text{s}$.
Steps to Solve
- Understanding gravitational acceleration on Earth and Moon
The gravitational acceleration on Earth is approximately $g_{Earth} \approx 9.81 , \text{m/s}^2$. The acceleration on the Moon, given it is $\frac{1}{6}$ that of Earth, is: $$ g_{Moon} = \frac{1}{6} g_{Earth} = \frac{9.81}{6} \approx 1.635 , \text{m/s}^2 $$
- Using maximum height to find initial velocity on Earth
The maximum height ($h$) achieved by a projectile can be calculated by the formula: $$ h = \frac{v^2}{2g} $$ Where $v$ is the initial velocity. Rearranging gives: $$ v = \sqrt{2gh} $$
For the height of 20 m on Earth: $$ v_{Earth} = \sqrt{2 \cdot 9.81 \cdot 20} $$
Calculating this gives: $$ v_{Earth} = \sqrt{392.4} \approx 19.8 , \text{m/s} $$
- Calculating maximum height on the Moon using the initial velocity
Using the same initial velocity on the Moon: $$ h_{Moon} = \frac{v_{Earth}^2}{2g_{Moon}} $$
Substituting the values: $$ h_{Moon} = \frac{(19.8)^2}{2 \cdot 1.635} $$
Calculating this results in: $$ h_{Moon} = \frac{392.04}{3.27} \approx 120.0 , \text{m} $$
- Calculating total time of flight on the Moon
The total time of flight ($T$) for a projectile is given by: $$ T = \frac{2v_{Earth}}{g_{Moon}} $$
Substituting the values: $$ T = \frac{2 \cdot 19.8}{1.635} $$
Calculating this gives: $$ T \approx \frac{39.6}{1.635} \approx 24.23 , \text{s} $$
- Finding total time from the throw until the baseball is caught
Since total time includes the ascent and descent, for our previous maximum height result, we need to clarify that:
- The time of ascent equals the time of descent, thus we divide by 2 for ascent:
Using the ascent only: $$ T_{ascent} = \frac{39.6}{2 \cdot 1.635} \approx 12.1 , \text{s} $$
On the Moon, you could throw the baseball to a maximum height of approximately $120.0 , \text{m}$, and the time elapsed from throw to catch would be approximately $24.23 , \text{s}$.
More Information
On the Moon, the lower gravitational force allows the baseball to reach a significantly higher altitude. The elongated time of flight showcases the Moon's weaker gravitational influence, extending how long the baseball stays in the air compared to Earth.
Tips
- Confusing the total time of flight with just the ascent time. Remember, the total time includes both ascent and descent.
- Miscalculating the gravitational acceleration on the Moon; ensure you take $\frac{1}{6}$ of Earth's gravity correctly.
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