Problem 59 - College Physics

Question

. A water hose is used to fill a large cylindrical storage tank of diameter \(D\) and height 2\(D\) The hose shoots the water at \(45^{\circ}\) above the horizontal from the same level as the base of the tank and is a distance 6\(D\) away (Fig. \(3.43 ) .\) For what range of launch speeds \(\left(v_{0}\right)\) will the water enter the tank? Ignore air resistance, and express your answer in terms of \(D\) and \(g .\)

Step-by-Step Solution

Answer

Projectile launch speeds range is \( \sqrt{\frac{13gD}{2}} \leq v_0 \leq 2\sqrt{13gD} \).

Step 1: Analyze the Problem

We need to find the range of launch speeds \( v_0 \) for which the water from the hose enters the cylindrical tank. We're given the tank's diameter \( D \) and height \( 2D \). The hose is placed 6\( D \) away from the tank, and the water is launched at a 45° angle. We are asked to use these parameters to compute \( v_0 \) in terms of \( D \) and \( g \).

Step 2: Break Down Projectile Motion into Components

Project the launch velocity \( v_0 \) into horizontal and vertical components. Use the angle 45°, which gives us: \[ v_{0x} = v_0 \cos{45^{\circ}} = \frac{v_0}{\sqrt{2}} \] \[ v_{0y} = v_0 \sin{45^{\circ}} = \frac{v_0}{\sqrt{2}} \]

Step 3: Find Horizontal Motion Equations

Use the horizontal component to solve for time \( t \) it takes for the water to travel the horizontal distance to the base of the tank, which is 6D + D/2 (the full horizontal distance from launcher to far edge of the tank): \[ x = v_{0x} \cdot t \Rightarrow 6D + \frac{D}{2} = \frac{v_0}{\sqrt{2}} \cdot t \] Solving for \( t \), we get: \[ t = \frac{13D}{2} \cdot \frac{\sqrt{2}}{v_0} \]

Step 4: Solve Vertical Motion for Entrance Range

Now, use the vertical motion equation to find when the water is within the height of the tank. The vertical position at time \( t \) is: \[ y = v_{0y}t - \frac{1}{2}gt^2 = \frac{v_0}{\sqrt{2}} \left(\frac{13D}{2} \cdot \frac{\sqrt{2}}{v_0}\right) - \frac{1}{2}g\left(\frac{13D}{2} \cdot \frac{\sqrt{2}}{v_0}\right)^2 \] Simplify this to ensure \(0 \leq y \leq 2D\).

Step 5: Calculate the Range of Launch Speeds

Set up the inequalities from Step 4, ensuring that the water reaches from the base to the full height of the tank. Solving for \( v_0 \) gives us two inequalities to satisfy:\[ \frac{13D}{2} \leq v_0^2/g \leq \frac{4 \cdot 13D}{2} \]Therefore, solve for \( v_0 \): \[ \sqrt{\frac{13gD}{2}} \leq v_0 \leq \sqrt{2 \cdot 13gD} \] Thus, the range for \( v_0 \) is: \[ 2\sqrt{\frac{13gD}{2}} \] and above for complete clearance.

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. In our problem with the water hose filling the tank, kinematics helps us understand how the water travels through the air as a projectile. By observing the motion of the water, we can break down its path into a more manageable mathematical form.
When analyzing projectile motion, kinematics allows us to separate the motion into horizontal and vertical components. This separation simplifies the complex movement into two linear motions that are easier to work with. We use kinematic equations to predict how long it takes for the projectile to reach its target and whether it can hit the target at all.

Physics Problem Solving

Problem-solving in physics involves applying core principles to find answers to real-world scenarios. With our water hose problem, we approach it by identifying what we know and what we need to find out. We start with the facts — the distance of the hose from the tank, the launch angle, and the sizes of the tank.
The next step is to translate these details into mathematical terms using equations of motion and trigonometric identities. Here, solving the problem involves examining how the water's velocity breaks into horizontal and vertical components, then finding formulas that will give us the range of permissible launch speeds.
This methodical approach takes a complex question and breaks it into digestible parts, making it easier to reach a solution. By applying a systematic process, we can solve a wide variety of problems by adjusting the variables involved.

Trigonometric Angles

Trigonometric angles are pivotal when analyzing projectile motion because they allow us to decompose forces and velocities into two perpendicular components. For the water hose problem, we work with an angle of 45 degrees. This is a special angle where the sine and cosine values are equal, simplifying calculations.
With trigonometric functions, we split the initial velocity, represented by the launch speed of the water, into horizontal and vertical components. The formulas we use are:

• Horizontal component: \( v_{0x} = v_0 \cos{45^{\circ}} = \frac{v_0}{\sqrt{2}} \)
• Vertical component: \( v_{0y} = v_0 \sin{45^{\circ}} = \frac{v_0}{\sqrt{2}} \)

The knowledge of trigonometry hence allows us to compute how the initial speed of the projectile affects its range and height, which are critical in determining how and if the water will land inside the tank.

Vertical and Horizontal Components

Analyzing the motion into vertical and horizontal components is essential in tackling projectile problems. The key to solving these kinds of problems is breaking down the launch velocity into these two perpendicular motions.
For horizontal motion in our scenario, we identify how far the water travels before entering the tank. By using the equation for horizontal distance,\[ x = v_{0x} \cdot t \], we determine the time it takes for the water to reach the tank. This helps in specifying how the particles move in space.
For vertical motion, it's the vertical component of the velocity that dictates the trajectory's peak and the time the projectile sustains in the air. The formula \( y = v_{0y} t - \frac{1}{2}gt^2 \) lets us ensure the path aligns with the tank's height. Together, these components ensure a comprehensive view of the projectile's path and are essential in finding the valid range for the launch velocity.