Question
A jet of water squirts out horizontally from a hole on the side of the tank as shown below. If the hole has a diameter of 3.37 mm , what is the height of the water above the hole in the tank? Answer in units of cm. (Height of hole= 0.99 m) (Distance water squirts out to= 0.677 m)
Answers
GPT-4o mini
To determine the height of the water above the hole in the tank, we can use principles from fluid dynamics, particularly Torricelli's law. This states that the speed of efflux of fluid under the force of gravity through an orifice is given by:
\[
v = \sqrt{2gh}
\]
where:
- \( v \) is the velocity of the jet of water exiting the hole,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height of the water column above the hole.
To find \( h \), we also need to analyze the trajectory of the water as it squirts out. Recognizing that the water follows a projectile motion after exiting the hole, we can use kinematic equations.
First, we can calculate the time \( t \) it takes for the water to fall to the ground from the height of the hole, influenced by gravity:
Given:
- Height from which the water falls \( H = 0.99 \, \text{m} \)
- Horizontal distance traveled \( R = 0.677 \, \text{m} \)
The vertical motion of the water can be described as follows. The distance fallen (which is the height of the hole \( H \)) is given by:
\[
H = \frac{1}{2} g t^2
\]
Rearranging this gives:
\[
t = \sqrt{\frac{2H}{g}}
\]
Now, substituting in the values:
\[
t = \sqrt{\frac{2 \times 0.99 \, \text{m}}{9.81 \, \text{m/s}^2}} \\
t \approx \sqrt{\frac{1.98}{9.81}} \\
t \approx \sqrt{0.202} \\
t \approx 0.449 \, \text{s}
\]
Next, we can find the horizontal velocity \( v \) of the water exiting the hole using:
\[
R = v t \Rightarrow v = \frac{R}{t}
\]
Substituting the values we have:
\[
v = \frac{0.677 \, \text{m}}{0.449 \, \text{s}} \approx 1.505 \, \text{m/s}
\]
Now that we have \( v \), we can go back to Torricelli's equation to find \( h \):
\[
v = \sqrt{2gh} \\
h = \frac{v^2}{2g}
\]
Substituting \( v \) and \( g \):
\[
h = \frac{(1.505 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \\
h = \frac{2.260}{19.62} \approx 0.115 \, \text{m}
\]
Converting this to centimeters:
\[
h \approx 11.5 \, \text{cm}
\]
Finally, to find the total height of the water above the hole, we need to add the height of the hole itself (the distance from the ground to the hole) to our calculated height \( h \):
Thus, the height of the water above the hole in the tank is:
\[
H_{total} = h + \text{height of the hole} = 11.5 \, \text{cm} + 99 \, \text{cm} \\
H_{total} = 110.5 \, \text{cm}
\]
Therefore, the final answer for the height of water above the hole in the tank is approximately:
\[
\boxed{110.5} \, \text{cm}
\]
\[
v = \sqrt{2gh}
\]
where:
- \( v \) is the velocity of the jet of water exiting the hole,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height of the water column above the hole.
To find \( h \), we also need to analyze the trajectory of the water as it squirts out. Recognizing that the water follows a projectile motion after exiting the hole, we can use kinematic equations.
First, we can calculate the time \( t \) it takes for the water to fall to the ground from the height of the hole, influenced by gravity:
Given:
- Height from which the water falls \( H = 0.99 \, \text{m} \)
- Horizontal distance traveled \( R = 0.677 \, \text{m} \)
The vertical motion of the water can be described as follows. The distance fallen (which is the height of the hole \( H \)) is given by:
\[
H = \frac{1}{2} g t^2
\]
Rearranging this gives:
\[
t = \sqrt{\frac{2H}{g}}
\]
Now, substituting in the values:
\[
t = \sqrt{\frac{2 \times 0.99 \, \text{m}}{9.81 \, \text{m/s}^2}} \\
t \approx \sqrt{\frac{1.98}{9.81}} \\
t \approx \sqrt{0.202} \\
t \approx 0.449 \, \text{s}
\]
Next, we can find the horizontal velocity \( v \) of the water exiting the hole using:
\[
R = v t \Rightarrow v = \frac{R}{t}
\]
Substituting the values we have:
\[
v = \frac{0.677 \, \text{m}}{0.449 \, \text{s}} \approx 1.505 \, \text{m/s}
\]
Now that we have \( v \), we can go back to Torricelli's equation to find \( h \):
\[
v = \sqrt{2gh} \\
h = \frac{v^2}{2g}
\]
Substituting \( v \) and \( g \):
\[
h = \frac{(1.505 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \\
h = \frac{2.260}{19.62} \approx 0.115 \, \text{m}
\]
Converting this to centimeters:
\[
h \approx 11.5 \, \text{cm}
\]
Finally, to find the total height of the water above the hole, we need to add the height of the hole itself (the distance from the ground to the hole) to our calculated height \( h \):
Thus, the height of the water above the hole in the tank is:
\[
H_{total} = h + \text{height of the hole} = 11.5 \, \text{cm} + 99 \, \text{cm} \\
H_{total} = 110.5 \, \text{cm}
\]
Therefore, the final answer for the height of water above the hole in the tank is approximately:
\[
\boxed{110.5} \, \text{cm}
\]