Question
A student drops a rock over the edge of the well and hears it splash into water after 3 seconds. Write a function in the form h(t)=-16t^2+vsubcscript0t+hsubscript0 to determine the height of the rock above the bottom of the well t seconds after the student drops the rock. The total height of the well is 200ft. What is the distance from the surface of the water to the bottom of the well?
Answers
GPT-4o mini
To solve the problem, we need to determine the height function of the rock as it falls and also understand the relationship between the splashing sound and the time it takes for the rock to reach the water.
1. **Understanding the problem:**
- The total height of the well is 200 feet.
- The rock is dropped, so its initial velocity (\(v_0\)) is 0.
- The function \(h(t)\) describes the height of the rock at time \(t\):
\[
h(t) = -16t^2 + v_0 t + h_0
\]
Since the rock is dropped, we have \(v_0 = 0\), and \(h_0\) is the initial height (200 feet):
\[
h(t) = -16t^2 + 200
\]
2. **Finding when the rock reaches the water:**
- The rock reaches the water when \(h(t) = 0\):
\[
0 = -16t^2 + 200
\]
Rearranging gives:
\[
16t^2 = 200 \\
t^2 = \frac{200}{16} = 12.5 \\
t = \sqrt{12.5} \approx 3.54 \text{ seconds}
\]
3. **Total time heard by student:**
- The student hears the splash after 3 seconds.
- This means the total time taken \(T\) for the rock to fall and the sound to travel back up is 3 seconds.
- Therefore, the time it takes for the rock to fall to the water (let's call this \(t_r\)) plus the time it takes for the sound to travel back up should equal 3 seconds:
\[
t_r + t_s = 3
\]
Where \(t_s\) is the time for sound to travel back up.
4. **Speed of sound:**
- The speed of sound in air is approximately \(343 \text{ m/s}\) or about \(1125 \text{ ft/s}\).
- The distance sound travels is the height of the well, which is 200 feet. Therefore, the time taken for the sound to travel back up (\(t_s\)) is:
\[
t_s = \frac{200}{1125} \approx 0.178 \text{ seconds}
\]
5. **Calculating \(t_r\):**
- Using our earlier equation:
\[
t_r + 0.178 = 3 \\
t_r = 3 - 0.178 \approx 2.822 \text{ seconds}
\]
Now, we also want to find the distance from the surface of the water to the bottom of the well (which is actually the height of the water itself).
6. **Finding height of the water:**
- Since we calculated the time it takes for the rock to fall (2.822 seconds), we can find the height from the bottom of the well to the surface of the water:
\[
h(t_r) = -16(2.822)^2 + 200
\]
\[
h(2.822) = -16(7.951) + 200 \\
h(2.822) \approx -127.216 + 200 \approx 72.784 \text{ feet}
\]
- Therefore, the distance from the surface of the water to the bottom of the well is approximately **72.784 feet**.
In summary, the height function for the rock is:
\[
h(t) = -16t^2 + 200
\]
And the distance from the surface of the water to the bottom of the well is approximately **72.784 feet**.
1. **Understanding the problem:**
- The total height of the well is 200 feet.
- The rock is dropped, so its initial velocity (\(v_0\)) is 0.
- The function \(h(t)\) describes the height of the rock at time \(t\):
\[
h(t) = -16t^2 + v_0 t + h_0
\]
Since the rock is dropped, we have \(v_0 = 0\), and \(h_0\) is the initial height (200 feet):
\[
h(t) = -16t^2 + 200
\]
2. **Finding when the rock reaches the water:**
- The rock reaches the water when \(h(t) = 0\):
\[
0 = -16t^2 + 200
\]
Rearranging gives:
\[
16t^2 = 200 \\
t^2 = \frac{200}{16} = 12.5 \\
t = \sqrt{12.5} \approx 3.54 \text{ seconds}
\]
3. **Total time heard by student:**
- The student hears the splash after 3 seconds.
- This means the total time taken \(T\) for the rock to fall and the sound to travel back up is 3 seconds.
- Therefore, the time it takes for the rock to fall to the water (let's call this \(t_r\)) plus the time it takes for the sound to travel back up should equal 3 seconds:
\[
t_r + t_s = 3
\]
Where \(t_s\) is the time for sound to travel back up.
4. **Speed of sound:**
- The speed of sound in air is approximately \(343 \text{ m/s}\) or about \(1125 \text{ ft/s}\).
- The distance sound travels is the height of the well, which is 200 feet. Therefore, the time taken for the sound to travel back up (\(t_s\)) is:
\[
t_s = \frac{200}{1125} \approx 0.178 \text{ seconds}
\]
5. **Calculating \(t_r\):**
- Using our earlier equation:
\[
t_r + 0.178 = 3 \\
t_r = 3 - 0.178 \approx 2.822 \text{ seconds}
\]
Now, we also want to find the distance from the surface of the water to the bottom of the well (which is actually the height of the water itself).
6. **Finding height of the water:**
- Since we calculated the time it takes for the rock to fall (2.822 seconds), we can find the height from the bottom of the well to the surface of the water:
\[
h(t_r) = -16(2.822)^2 + 200
\]
\[
h(2.822) = -16(7.951) + 200 \\
h(2.822) \approx -127.216 + 200 \approx 72.784 \text{ feet}
\]
- Therefore, the distance from the surface of the water to the bottom of the well is approximately **72.784 feet**.
In summary, the height function for the rock is:
\[
h(t) = -16t^2 + 200
\]
And the distance from the surface of the water to the bottom of the well is approximately **72.784 feet**.