Problem 68 - University Physics with Modern Physics

Question

A firefighting crew uses a water cannon that shoots water at 25.0 m/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 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.)

Step-by-Step Solution

Answer

The cannon can be positioned at either 8.85 meters or 51.6 meters from the building.

Step 1: Initial Setup

Before starting the calculations, sketch a trajectory diagram of the water to help visualize the path. The trajectory is a parabola because the angle is above the horizontal due to projectile motion. This will aid in distinguishing the two points of height that the water can reach before descending, meeting the target value of 10.0 m.

Step 2: Determine Horizontal and Vertical Components of Velocity

Decompose the initial velocity of 25.0 m/s into horizontal and vertical components. The horizontal component is \(v_{x} = 25.0 \cos 53.0^{\circ}\) and the vertical component is \(v_{y} = 25.0 \sin 53.0^{\circ}\). Calculating these, we get \(v_{x} \approx 15.0 \text{ m/s}\) and \(v_{y} \approx 20.0 \text{ m/s}\).

Step 3: Write the Equation of Motion for Vertical Displacement

The vertical motion equation considering the initial velocity, gravity, and time is: \(y = v_{y}t - \frac{1}{2}gt^2\). Set \(y = 10.0\) m and solve for time \(t\), using \(g = 9.8\text{ m/s}^2\). This gives us a quadratic equation: \(10 = 20t - 4.9t^2\).

Step 4: Solve the Quadratic Equation for Time

Solve the quadratic equation \(4.9t^2 - 20t + 10 = 0\) using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4.9\), \(b = -20\), and \(c = 10\). This yields two solutions for time: \(t_1 \approx 0.59\text{ s} \) and \(t_2 \approx 3.44\text{ s}\).

Step 5: Calculate Horizontal Distances

Using the calculated times, determine the horizontal distances using the formula: \(x = v_{x}t\). For \(t_1 = 0.59\text{ s}\), \(x_1 = 15.0 \times 0.59 \approx 8.85\text{ m}\). For \(t_2 = 3.44\text{ s}\), \(x_2 = 15.0 \times 3.44 \approx 51.6\text{ m}\).

Step 6: Conclusion

Based on the solutions for the times, the firefighting crew can position their cannon either 8.85 meters or 51.6 meters away from the building to hit the target height of 10.0 meters with the water stream.

Key Concepts

horizontal and vertical components

In projectile motion, understanding the horizontal and vertical components is key to solving any related problems. Imagine you have a water cannon like in the exercise. It shoots water at a certain speed and angle. This speed, known as initial velocity, can be broken down into two parts, horizontal and vertical, using trigonometry.
The horizontal component (\( v_x \)) is responsible for the forward movement of the projectile. It can be found using the cosine of the angle with the formula: \( v_x = v_{initial} \cos{\theta} \). The vertical component (\( v_y \)), however, affects how high the projectile will go and is calculated with the sine of the angle: \( v_y = v_{initial} \sin{\theta} \).

• For example, with a velocity of 25 m/s at a 53-degree angle, the horizontal component is about 15 m/s.
• The vertical component is around 20 m/s.

These decomposed values help us understand and predict the motion path, breaking it into manageable sections.

trajectory of a projectile

Once we know our horizontal and vertical components, we can predict the path or trajectory of the projectile — in this case, the water stream. In physics, this path forms a parabolic shape due to gravity's pull.
The initial speed sends the projectile upward, but gravity slows it down until it stops for a brief moment at its highest point. Then, it pulls the projectile back down.
Visualizing the trajectory helps in planning where the water stream needs to land. The exercise above attempts to find out where to position the cannon, ensuring the water reaches a precise height. By sketching a diagram, the arc of the water can be seen clearly.

• This helps firefighters determine two points at height 10m where the stream will pass through: one on the way up, and another on the way down.

This parabolic path highlights how both angles of launch and speed influence the outcome of projectile motion.

quadratic equation in physics

Quadratic equations are mathematical representations often used to solve physics problems involving projectile motion. In the exercise, the vertical position is described by a quadratic equation, highlighting how changes in time affect the projectile's height.
The formula used is \( y = v_y t - \frac{1}{2}gt^2 \). We know our target height is 10 m. By substituting known values, we find \( 10 = 20t - 4.9t^2 \). This forms a quadratic equation, \( 4.9t^2 - 20t + 10 = 0 \), that can be solved using the quadratic formula.
Solving it allows us to find two different times when the projectile reaches this height:

• \( t_1 \approx 0.59 \ ext{ s} \)
• \( t_2 \approx 3.44 \ ext{ s} \)

These solutions correspond to two possible distances from the building where the cannon can be positioned. The quadratic nature of the equation reflects the reality of motion influenced by gravity, displaying both ascend and descend points within the trajectory.