Question
If a rock falls for 13 seconds how far up was it dropped from
Answers
GPT-4o mini
To calculate the distance a rock falls when dropped from a certain height, you can use the following kinematic equation for uniformly accelerated motion due to gravity:
\[
d = \frac{1}{2} g t^2
\]
where:
- \( d \) is the distance fallen,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) near the surface of the Earth),
- \( t \) is the time in seconds.
For a rock falling for 13 seconds:
\[
d = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (13 \, \text{s})^2
\]
Calculating this step by step:
1. Calculate \( (13 \, \text{s})^2 = 169 \, \text{s}^2 \).
2. Now multiply by \( 9.81 \, \text{m/s}^2 \):
\[
9.81 \, \text{m/s}^2 \times 169 \, \text{s}^2 \approx 1655.49 \, \text{m}
\]
3. Finally, multiply by \( \frac{1}{2} \):
\[
d \approx \frac{1}{2} \times 1655.49 \, \text{m} \approx 827.74 \, \text{m}
\]
Thus, the rock was dropped from approximately **827.74 meters**.
\[
d = \frac{1}{2} g t^2
\]
where:
- \( d \) is the distance fallen,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) near the surface of the Earth),
- \( t \) is the time in seconds.
For a rock falling for 13 seconds:
\[
d = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (13 \, \text{s})^2
\]
Calculating this step by step:
1. Calculate \( (13 \, \text{s})^2 = 169 \, \text{s}^2 \).
2. Now multiply by \( 9.81 \, \text{m/s}^2 \):
\[
9.81 \, \text{m/s}^2 \times 169 \, \text{s}^2 \approx 1655.49 \, \text{m}
\]
3. Finally, multiply by \( \frac{1}{2} \):
\[
d \approx \frac{1}{2} \times 1655.49 \, \text{m} \approx 827.74 \, \text{m}
\]
Thus, the rock was dropped from approximately **827.74 meters**.