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Chapter 3: Problem 55

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 two possibilities; can you get them both? (Hint: Start with a sketch showing the trajectory of the water.)

Short Answer

Expert verified
About 8.1 m and 23.6 m are the two possible distances.

Step by step solution

01

Break Down the Problem

We have a projectile motion problem where water is fired at an initial speed of 25.0 m/s and an angle of 53.0 degrees. The goal is to find the horizontal distance (range) needed to reach a height of 10.0 m.
02

Resolve Initial Velocity into Components

Calculate the initial velocity components. Use sine and cosine to resolve the initial velocity:\( v_{0x} = v_0 \cos \theta = 25.0 \cos 53.0^{\circ} \)\( v_{0y} = v_0 \sin \theta = 25.0 \sin 53.0^{\circ} \)
03

Use Vertical Motion to Find Time

For vertical motion, use the equation:\( y = v_{0y}t - \frac{1}{2}gt^2 \)Set \( y = 10.0 \text{ m} \) and solve for time \( t \) using \( 10.0 = (25.0 \sin 53.0^{\circ})t - \frac{1}{2}(9.8)t^2 \)
04

Solve the Quadratic Equation

Rearrange the equation from Step 3:\( 4.9t^2 - 19.94t + 10 = 0 \)Use the quadratic formula:\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)Plug in \( a = 4.9 \), \( b = -19.94 \), \( c = 10 \). This yields two times.
05

Calculate Horizontal Distance for Each Time

For each time solution from the quadratic equation, calculate the horizontal distance using:\( x = v_{0x}t \) For both positive time results, calculate:\( x = (25.0 \cos 53.0^{\circ})t \)
06

Choose the Valid Distances

Calculate the horizontal distances from Step 5 and identify both positive values. These values will give us the two possible distances where the water can reach a height of 10.0 m.

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Key Concepts

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

Initial Velocity Components
When dealing with projectile motion problems, one of the first steps is to break down the initial velocity into components. This helps in analyzing the motion in both the horizontal and vertical directions separately.

In this case, we have an initial speed of 25.0 m/s at an angle of 53.0 degrees. By using trigonometric functions such as sine and cosine, we can resolve this velocity into two components:
  • Horizontal Velocity, \( v_{0x} = v_0 \cos \theta \). For the given example:
    \( v_{0x} = 25.0 \cos 53.0^{\circ} \).
  • Vertical Velocity, \( v_{0y} = v_0 \sin \theta \). Here:
    \( v_{0y} = 25.0 \sin 53.0^{\circ} \).
By separating these components, we can work on both the vertical and horizontal aspects of the motion independently, making calculations easier.
Quadratic Equation
Once we break down the components, we often use the quadratic equation to solve for the time 't' at which the projectile reaches a certain height.

The vertical motion is given by the equation:\[ y = v_{0y}t - \frac{1}{2}gt^2 \]In this problem, we set \( y = 10.0 \) m to find when the projectile reaches the blaze. Substituting the known values will give us a quadratic equation:\[ 4.9t^2 - 19.94t + 10 = 0 \]

This is a standard form of a quadratic equation, \( at^2 + bt + c = 0 \). Here:
  • \( a = 4.9 \)
  • \( b = -19.94 \)
  • \( c = 10 \)
The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) will give us two possible times which the water cannon will be at the required height.
Trajectory
A trajectory is the path that a projectile follows through space as a function of time.Understanding trajectory is essential for predicting where a projectile will land.In our example, the water follows a parabolic path due to the combined effects of horizontal and vertical motions.

The shape of this path is determined by:
  • Initial velocity - both its magnitude and direction.
  • Acceleration due to gravity - acting downwards at 9.8 m/s\(^2\).
To visualize this, draw a curve starting from the point of launch that rises, reaches a peak, and then descends.Knowing the trajectory allows us to calculate important values like range and maximum height.
Vertical Motion Equations
Vertical motion equations describe how the vertical position of the projectile changes over time due to gravity. The equation used in this scenario is:\[ y = v_{0y}t - \frac{1}{2}gt^2 \]

Breaking this down:
  • \(y\): Vertical position (height).
  • \(v_{0y}\): Initial vertical velocity.
  • \(t\): Time.
  • \(g\): Gravitational constant (9.8 m/s\(^2\)).
This equation accounts for the initial upward motion and the subsequent downward motion due to gravity.It helps to determine at what time the projectile reaches a specified height, crucial for our firefighting crew to position the water cannon accurately.

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Most popular questions from this chapter

A \(\mathbf{A}\) world record. In the shot put, a standard track-and- field event, a 7.3 \(\mathrm{kg}\) object (the shot) is thrown by releasing it at approximately \(40^{\circ}\) over a straight left leg. The world record for distance, set by Randy Barnes in \(1990,\) is 23.11 \(\mathrm{m} .\) Assuming that Barnes released the shot put at \(40.0^{\circ}\) from a height of 2.00 \(\mathrm{m}\) above the ground, with what speed, in \(\mathrm{m} / \mathrm{s}\) and mph, did he release it?

A military helicopter on a training mission is flying horizontally at a speed of 60.0\(\mathrm{m} / \mathrm{s}\) when it accidentally drops a bomb(fortunately, not armed) at an elevation of 300 \(\mathrm{m.}\) You can ignore air resistance. (a) How much time is required for the bomb to reach the earth? (b) How far does it travel horizontally while falling? (c) Find the horizontal and vertical components of the bomb's velocity just before it strikes the earth. (d) Draw graphs of the horizontal distance vs. time and the vertical distance vs. time for the bomb's motion. (e) If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?

. A certain cannon with a fixed angle of projection has a range of 1500 \(\mathrm{m}\) . What will be its range if you add more powder so that the initial speed of the cannonball is tripled?

\(\bullet\) A wall clock has a second hand 15.0 \(\mathrm{cm}\) long. What is the radial acceleration of the tip of this hand?

If the time of flight of the ball is \(t\) seconds, at what point in time will the ball have zero vertical velocity? A. \(t / 4 \mathrm{s}\) B. \(t / 4 \mathrm{s}\) C. \(t / 2 \mathrm{s}\) D. \(t / 2 \mathrm{s}\) E. There is no place on the path where the vertical velocity is zero.
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