Problem 20 - University Physics with Modern Physics

Question

Firemen use a high-pressure hose to shoot a stream of water at a burning building. The water has a speed of 25.0 m/s as it leaves the end of the hose and then exhibits projectile motion. The firemen adjust the angle of elevation \(\alpha\) of the hose until the water takes 3.00 s to reach a building 45.0 m away. Ignore air resistance; assume that the end of the hose is at ground level. (a) Find \(\alpha\). (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

Step-by-Step Solution

Answer

(a) \(\alpha \approx 53.13^\circ\). (b) Speed is 15 m/s; acceleration is -9.81 m/s². (c) 15.86 m high, speed \(\approx 17.5\) m/s.

Step 1: Breakdown of Horizontal Motion

Since the water reaches the building in 3.0 seconds, and the horizontal distance to the building is 45.0 meters, we can calculate the horizontal component of the velocity. The horizontal velocity is given by:\[ v_{x} = \frac{d}{t} = \frac{45.0}{3.0} = 15.0 \, \text{m/s} \]

Step 2: Calculate the Angle of Elevation \(\alpha\)

The known initial speed is 25.0 m/s. Using trigonometric identities, we relate the horizontal speed \(v_{x}\) to the initial speed and angle of elevation:\[ v_{x} = v_{0} \cos \alpha = 25.0 \cos \alpha \]Thus, we have:\[ 25.0 \cos \alpha = 15.0 \]Solving for \(\alpha\):\[ \cos \alpha = \frac{15.0}{25.0} = 0.6 \]\[ \alpha = \cos^{-1}(0.6) \approx 53.13^\circ \]

Step 3: Calculate Vertical Motion at the Highest Point

At the highest point, the vertical component of the velocity is zero. Thus, the speed at the highest point is equal to the horizontal component of the speed, i.e., 15.0 m/s. The acceleration is always \(-9.81 \, \text{m/s}^{2}\) due to gravity, acting downwards.

Step 4: Determine the Height Struck on the Building

The vertical component of the initial velocity is:\[ v_{y} = v_{0} \sin \alpha = 25.0 \sin(53.13^\circ) \approx 20.0 \, \text{m/s} \]Using the equation of motion:\[ h = v_{y} t - \frac{1}{2} g t^2 \]\[ h = 20.0 \times 3.0 - \frac{1}{2} \times 9.81 \times (3.0)^2 \]\[ h = 60.0 - 44.145 \approx 15.86 \, \text{m} \]

Step 5: Calculate the Speed Just Before Hitting the Building

The vertical speed component just before hitting the building, using final vertical velocity formula:\[ v_{yf} = v_{y} - g \cdot t = 20.0 - 9.81 \times 3.0 \approx -9.43 \, \text{m/s} \]Magnitude of total speed just before impact:\[ v = \sqrt{v_{x}^2 + v_{yf}^2} = \sqrt{15.0^2 + (-9.43)^2} \]\[ v \approx 17.5 \, \text{m/s} \]

Key Concepts

Horizontal Velocity

In projectile motion, horizontal velocity remains constant because there are no horizontal forces acting on the object (assuming air resistance is negligible). This means the horizontal component of a projectile's velocity does not change during its flight. In this specific exercise, the water reaches the building in 3.0 seconds, and covers a horizontal distance of 45.0 meters. By using the formula for horizontal velocity:

• \( v_x = \frac{d}{t} \)
• Where \( d \) is the distance and \( t \) is the time.

We find that:

• \( v_x = \frac{45.0}{3.0} = 15.0 \, \text{m/s} \)

This constant horizontal velocity is a critical concept in understanding projectile motion as it simplifies calculations of distance and time.

Vertical Velocity

Unlike horizontal velocity, the vertical component of a projectile's motion is affected by gravity, which causes a constant acceleration of \( -9.81 \, \text{m/s}^2 \).In the initial launch, the vertical velocity is calculated using the formula:

• \( v_y = v_0 \sin \alpha \)
• Where \( v_0 \) is the initial speed and \( \alpha \) is the angle of elevation.

For this problem, after calculating the angle (\( \alpha \approx 53.13^\circ \)), we use the equation to find:

• \( v_y = 25.0 \sin(53.13^\circ) \approx 20.0 \, \text{m/s} \)

At the highest point of its trajectory, the vertical velocity decreases to zero because the upward and downward velocities cancel out. As the water descends, the vertical velocity increases negatively due to gravity.

Angle of Elevation

The angle of elevation is crucial in determining the trajectory and range of the projectile. It is the angle between the initial velocity vector and the horizontal axis. To find this angle, we use the relationship between horizontal velocity \( v_x \) and initial speed \( v_0 \):

• \( v_x = v_0 \cos \alpha \)

Given the horizontal velocity \( v_x = 15.0 \, \text{m/s} \) and \( v_0 = 25.0 \, \text{m/s} \), solving for the angle yields:

• \( \cos \alpha = \frac{15.0}{25.0} = 0.6 \)
• \( \alpha = \cos^{-1}(0.6) \approx 53.13^\circ \)

This angle ensures the water reaches the necessary height and distance to strike the building. Adjusting \( \alpha \) in real scenarios is how firemen achieve precise control over the stream's path.

Acceleration Due to Gravity

Acceleration due to gravity is the force responsible for changing the vertical component of a projectile's velocity. It is a constant \( 9.81 \, \text{m/s}^2 \) acting downwards, affecting only the vertical motionThis consistent acceleration means at every second, the vertical velocity changes by \( -9.81 \, \text{m/s} \).For the water in this exercise, when it reaches its highest point, the net vertical speed is zero before gravity accelerates the water down again. This downward acceleration affects how long the water stays in the air and at what speed it hits the building. This can be calculated using the formula for final velocity in free-fall:

• \( v_{yf} = v_y - g \cdot t \)

Which leads to:

• \( v_{yf} \approx -9.43 \, \text{m/s} \)

Understanding this concept is vital as it shows how objects will always accelerate towards Earth at a predictable rate.