A jet of water squirts out horizontally froma hole on the side of the tank as shown below. If the hole has a diameter of 3.62 mm, whatis the height of the water above the hole inthe tank?Answer in units of cm.

A jet of water squirts out horizontally froma hole on the side of the tank as shown below If the hole has a diameter of 362 mm whatis the height of the water ab class=

Respuesta :

The jet of water emerges with a speed v horizontally. Then, it falls a vertical distance of 1.58, and during the time of its fall until it hits the ground, it travels a distance of 0.586 meters.

Since the water moves according to the Projectile Motion formulas, the horizontal motion is uniform and the vertical motion is uniformly accelerated from rest.

Then, we can use the vertical distance to find the time that it takes for the water to reach the ground. Then, divide the horizontal distance traveled by the water by the time it took for the water to fall to find the horizontal speed of the water.

Finally, use Torricelli's Theorem to relate the height of the water above the hole in the pipe to the horizontal speed. Find the height of the hole by isolating it and replacing the value of the known variables.

Step 1: Time

Since the water falls from rest a distance of 1.58m, and the time it takes for an object to fall a distance d in free fall from rest is given by:

[tex]t=\sqrt[]{\frac{2d}{g}}[/tex]

Then, replace d=1.58m and the value of the acceleration of gravity g=9.81m/s^2 to find the time it takes for the water to fall:

[tex]t=\sqrt[]{\frac{2(1.58m)}{9.81\frac{m}{s^2}}}=0.5675564161\ldots s[/tex]

Step 2: Speed

Since the water travels a horizontal distance of 0.586m during that time, then, the speed of the water jet as it emerges from the hole, is:

[tex]v=\frac{x}{t}=\frac{0.586m}{0.5675564161\ldots s}=1.032496477\ldots\frac{m}{s}[/tex]

Step 3: Height (Torricelli's Theorem)

According to Torricelli's Theorem, the speed of a liquid jet emerging from a hole located a distance h below the level of water of a large container, is:

[tex]v=\sqrt[]{2gh}[/tex]

Isolate h from the equation:

[tex]h=\frac{v^2}{2g}[/tex]

Replace v=1.032496477...m/s and g=9.81m/s^2 to find the height of the water above the hole in the tank:

[tex]h=\frac{(1.032406477\ldots\frac{m}{s})^2}{2(9.81\frac{m}{s^2})}=0.05433481013\ldots m\approx5.43\times10^{-2\text{ }}m[/tex]

Write the answer in cm:

[tex]h=5.43cm[/tex]

Therefore, the height of the water above the hole in the tank is 5.43 centimeters.