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Chapter 3: Problem 66

A firefighting crew uses a water cannon that shoots water at \(25.0 \mathrm{~m} / \mathrm{s}\) at a fixed angle of \(53.0^{\circ}\) above the horizontal. The firefighters want to direct the water at a blaze that is \(10.0 \mathrm{~m}\) above ground level. How far from the building should they position their cannon? There are \(t w o\) possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.

Short Answer

Expert verified
The firefighters can position their cannon either approximately 6.5 m or 20.9 m from the building.

Step by step solution

01

Identify the Variables

The initial speed of the water (\(V_i\)) is given as 25.0 m/s, the angle (\(胃\)) at which the water is shot is given as 53 degrees, and the distance above the ground (\(螖y\)) where the water needs to land is given as 10.0 m.
02

Separate into Vertical and Horizontal Components

The velocity of water has both horizontal and vertical components. These can be calculated as follows: \(V_{ix} = V_i \cdot cos(胃)\) for the horizontal component and \(V_{iy} = V_i \cdot sin(胃)\) for the vertical component.
03

Apply Equation of Motion to Find Time

Use the equation of motion to find the time it takes for the water to hit the blaze: \(螖y = V_{iy}t + 0.5gt^2\). Here, \(g\) is the acceleration due to gravity, which is -9.8 m/s^2 (negative because the upward direction is taken as positive). After rearranging, we will have a quadratic equation, which will yield two solutions as stated in the question.
04

Solve for Distance

Plug in the determined time(s) into the equation \(螖x = V_{ix} \cdot t\) to determine the two possible distances from the building that the firefighters can position their cannon.

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Key Concepts

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

Kinematic Equations
Kinematic equations are a set of equations that describe the motion of objects. They're particularly handy in analyzing projectile motion problems. In general, they help relate distances, velocities, accelerations, and time. When you understand how these factors interplay, you can predict anything about the motion of an object, be it horizontally launched or vertically boosted.
  • The first primary kinematic equation is: \[ 螖y = V_{iy}t + \frac{1}{2}at^2 \]where \(螖y\) is the vertical displacement, \(V_{iy}\) is the initial vertical velocity, \(a\) is acceleration, and \(t\) is time.
  • The second equation to remember is the horizontal motion, which has constant velocity as there's no horizontal acceleration: \[ 螖x = V_{ix}t \]where \(螖x\) is the horizontal displacement and \(V_{ix}\) is the initial horizontal velocity.
Understanding these equations allows us to dissect the complex aspects of projectile motion, making it simpler to analyze each part of the movement systematically.
Vertical and Horizontal Components
Every projectile is influenced by two types of motion: vertical and horizontal. By breaking down the velocity into its components, you're able to solve projectile motion problems more efficiently.
  • The vertical component (\[ V_{iy} \]) describes the upward or downward motion. It's determined by: \[ V_{iy} = V_i \cdot \sin(胃) \]
  • The horizontal component (\[ V_{ix} \]) accounts for motion parallel to the ground. It's calculated as: \[ V_{ix} = V_i \cdot \cos(胃) \]
By separately examining these components: The vertical motion tells us how high or low the projectile goes. The horizontal motion informs how far it travels. Together, these provide a full picture of the projectile's path.
Quadratic Equations
Quadratic equations pop up frequently in physics, especially when dealing with projectile motion problems involving time. The standard form is \[ at^2 + bt + c = 0 \]. Solving this requires either factoring, completing the square, or using the quadratic formula.
  • The quadratic formula is: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • The "\(\pm\)" indicates two potential solutions. In projectile motion, this often corresponds to two different times at which the projectile can reach a certain height.
This two-timing event reflects real-world scenarios where an object might rise past a certain point, and later fall back down through the same point. Mastering quadratic equations is essential for accurately predicting these events.
Problem Solving in Physics
Problem solving in physics doesn't just involve plugging numbers into formulas. It involves developing a thorough understanding of the problem first. Start with a sketch. Visualizing the scenario can clarify which components and equations are needed.
  • Identify the given variables and unknowns clearly. This helps you select the appropriate formulas.
  • Break down complex motion into simpler parts, such as vertical and horizontal components for projectiles.
  • Apply relevant kinematic equations for each component, solving for unknowns like time or distance.
  • Be prepared to interpret multiple solutions, as with quadratic results, and determine their real-world feasibility.
Developing these problem-solving skills transforms physics from intimidating to intuitive, allowing for a deeper understanding and quicker calculation efficiency.

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Most popular questions from this chapter

A roller coaster car moves in a vertical circle of radius \(R\). At the top of the circle the car has speed \(v_{1}\), and at the bottom of the circle it has speed \(v_{2},\) where \(v_{2}>v_{1} .\) (a) When the car is at the top of its circular path, what is the direction of its radial acceleration, \(a_{\mathrm{rad}, \text { top }} ?\) (b) When the car is at the bottom of its circular path, what is the direction of its radial acceleration, \(a_{\mathrm{rad}, \text { bottom }} ?\) (c) In terms of \(v_{1}\) and \(v_{2}\), what is the ratio \(a_{\text {rad, bottom }} / a_{\text {rad, top }} ?\)

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in \(2.70 \mathrm{~s},\) while Milada jumps horizontally with an initial speed of \(95.0 \mathrm{~cm} / \mathrm{s}\). How far from the base of the cliff will Milada hit the ground? Ignore air resistance.

Henrietta is jogging on the sidewalk at \(3.05 \mathrm{~m} / \mathrm{s}\) on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is \(38.0 \mathrm{~m}\) above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally \(9.00 \mathrm{~s}\) after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

A bird flies in the \(x y\) -plane with a velocity vector given by \(\overrightarrow{\boldsymbol{v}}=\left(\alpha-\beta t^{2}\right) \hat{\imath}+\gamma t \hat{\jmath},\) with \(\alpha=2.4 \mathrm{~m} / \mathrm{s}, \beta=1.6 \mathrm{~m} / \mathrm{s}^{3},\) and \(\gamma=4.0 \mathrm{~m} / \mathrm{s}^{2} .\) The positive \(y\) -direction is vertically upward. At \(t=0\) the bird is at the origin. (a) Calculate the position and acceleration vectors of the bird as functions of time. (b) What is the bird's altitude (y-coordinate) as it flies over \(x=0\) for the first time after \(t=0 ?\)

For this equipment to land at the front of the ship, at what a ship, which is moving at \(45.0 \mathrm{~cm} / \mathrm{s}\), before the ship can dock. This equipment is thrown at \(15.0 \mathrm{~m} / \mathrm{s}\) at \(60.0^{\circ}\) above the horizontal from the top of a tower at the edge of the water, \(8.75 \mathrm{~m}\) above the ship's deck (Fig. \(\mathbf{P 3 . 5 4}\) ). For this equipment to land at the front of the ship, at what distance \(D\) from the dock should the ship be when the equipment is thrown? Ignore air resistance.
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