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Chapter 12: Problem 71

You drill a small hole in the side of a vertical cylindrical water tank that is standing on the ground with its top open to the air. (a) If the water level has a height H, at what height above the base should you drill the hole for the water to reach its greatest distance from the base of the cylinder when it hits the ground? (b) What is the greatest distance the water will reach?

Short Answer

Expert verified
(a) Drill at \( h = \frac{H}{2} \). (b) Maximum distance is \( H \).

Step by step solution

01

Understand the Problem

We need to determine at what height above the base a hole should be drilled to maximize the horizontal distance the water travels when it exits the tank. This is a case of projectile motion where water exits through a hole due to pressure.
02

Set Up Formula for Horizontal Distance

The horizontal distance (R) the water travels can be determined by the formula for projectile motion: \[ R = v_x \times t \]where \( v_x \) is the horizontal velocity of the water and \( t \) is the time taken for the water to hit the ground.
03

Determine Exit Velocity

The velocity \( v_x \) with which the water exits the hole is given by Torricelli's theorem:\[ v_x = \sqrt{2g(H-h)} \]where \( g \) is the acceleration due to gravity, \( H \) is the initial water level height and \( h \) is the height of the hole above the base.
04

Calculate Time of Flight

The time \( t \) for the water to fall a vertical distance \( h \) (from hole to ground) is given by:\[ t = \sqrt{\frac{2h}{g}} \]
05

Substitute and Maximize Horizontal Distance

Substitute \( v_x \) and \( t \) into the equation for \( R \):\[ R = \sqrt{2g(H-h)} \times \sqrt{\frac{2h}{g}} = \sqrt{4h(H-h)} \]To maximize \( R \), differentiate \( R^2 = 4h(H-h) \) with respect to \( h \) and set the derivative equal to zero:\[ \frac{d}{dh}(4h(H-h)) = 4H - 8h = 0 \]Solving gives \( h = \frac{H}{2} \).
06

Calculate Greatest Distance

Substitute \( h = \frac{H}{2} \) back into the equation for \( R^2 \):\[ R^2 = 4\left(\frac{H}{2}\right)(H-\frac{H}{2}) = H^2 \]Thus, the greatest distance \( R \) is:\[ R = H \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torricelli's Theorem
Torricelli's Theorem is central to understanding the motion of a fluid exiting a container. It tells us how fast water will flow out of a hole in a tank. This theorem states that the speed of efflux of a fluid, under the force of gravity, is the same as if the fluid had fallen freely from the height of its source point to the exit point.
To put it simply, when you make a hole in the side of a tank full of water, the water flows out with a speed given by:\[ v_x = \sqrt{2g(H-h)} \]Here,
  • \( v_x \) is the horizontal velocity of the water as it exits the hole.
  • \( g \) is the acceleration due to gravity.
  • \( H \) is the initial height of the water level.
  • \( h \) is the height of the hole above the base of the tank.

This equation is derived from the principle of conservation of energy, where the potential energy is converted into kinetic energy.
Horizontal Distance
When talking about horizontal distance in projectile motion, we want to know how far an object will travel horizontally from its launch point. In this case of water from a tank, it exits a hole and travels a certain distance before hitting the ground.
The horizontal distance \( R \) can be calculated using the formula:\[ R = v_x \times t \]Where:
  • \( v_x \) is the horizontal velocity of the water as found using Torricelli's Theorem.
  • \( t \) is the time it takes for the water to hit the ground from the height of the hole.

Maximizing this horizontal distance involves finding the optimal height \( h \), for which the water can travel the furthest. By derivation, it was found that drilling the hole halfway up the height \( H/2 \) allows the water to travel the greatest distance before reaching the ground.
Time of Flight
Time of flight refers to how long the water is in motion from leaving the tank until it reaches the ground. This is a key factor in determining how far the water can travel horizontally.
To calculate the time \( t \) it takes for the water to fall from the hole to the ground, we use the equation:\[ t = \sqrt{\frac{2h}{g}} \]Where:
  • \( h \) is the vertical distance from the hole to the ground.
  • \( g \) is the constant acceleration due to gravity.

This time of flight is crucial in determining the horizontal distance because it affects how long the water has to travel horizontally before hitting the ground.
Gravity
Gravity plays a fundamental role in projectile motion by influencing both the speed and the trajectory of a projectile. It is the attractive force that pulls objects towards the Earth, causing objects to accelerate downwards.
When water flows out of the tank, gravity affects its vertical motion. In the given problem, gravity:
  • Determines the exit speed of the water through Torricelli's Theorem: The greater the gravitational pull, the faster the water exits.
  • Affects the time of flight: The constant \( g \) shows up in the time of flight formula indicating how long it takes for the water to reach the ground.

Understanding gravity is essential in predicting how far and fast the water will travel once it leaves the tank. In essence, without gravity, there wouldn't be the familiar parabolic trajectory seen in projectile motion.

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Most popular questions from this chapter

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill \(2200.355-\) L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the cross-sectional area is 8.00 \(\mathrm{cm}^{2} .\) At point \(1,1.35\) above point \(2,\) the cross-sectional area is 2.00 \(\mathrm{cm}^{2} .\) Find the (a) mass flow rate; (b) volume flow rate; (c) flow speeds at points 1 and \(2 ;\) (d) gauge pressure at point \(1 .\)

A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water completely fills the pipe. (a) At one point in the pipe the radius is 0.150 \(\mathrm{m}\) . What is the speed of the water at this point if water is flowing into this pipe at a steady rate of 1.20 \(\mathrm{m}^{3} / \mathrm{s} ?\) (b) At a second point in the pipe the water speed is 3.80 \(\mathrm{m} / \mathrm{s} .\) What is the radius of the pipe at this point?

A swimming pool is 5.0 \(\mathrm{m}\) long, 4.0 \(\mathrm{m}\) wide, and 3.0 \(\mathrm{m}\) deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h,\) and integrate this over the end of the pool.) Do not include the force due to air pressure.

A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 -kg woman to be able to stand on it without getting her feet wet?
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