Problem 71 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: Calculate Time of Flight

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

Step 5: 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 \( h = \frac{H}{2} \).

Step 6: 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:\[ R = H \].

Key Concepts

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.