Problem 23 - College Physics

Question

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.(a) Find the angle of elevation of the hose. (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 visit moving just before it hits the building?

Step-by-Step Solution

Answer

(a) Angle is about 53.13°. (b) Speed is 15 m/s, acceleration is -9.8 m/s². (c) Height is 15.9 m, speed is 17.7 m/s.

Step 1: Analyze the Horizontal Motion

The horizontal component of the velocity, \( v_{x} \), is given by \( v_{x} = v \cos(\theta) \), where \( v \) is the initial speed (25.0 m/s) and \( \theta \) is the angle of elevation. Using the formula for horizontal motion, \( x = v_{x} \cdot t \), where \( x = 45.0 \) m and \( t = 3.00 \) s, we have:\[45.0 = 25.0 \cos(\theta) \cdot 3.00.\]Solve for \( \cos(\theta) \):\[\cos(\theta) = \frac{45.0}{75.0} = 0.6.\]

Step 2: Determine the Angle of Elevation

Now that we have \( \cos(\theta) = 0.6 \), we can find the angle \( \theta \) by using the arccos function:\[\theta = \cos^{-1}(0.6).\]Using a calculator, \( \theta \approx 53.13^\circ \).

Step 3: Analyze the Vertical Motion at the Highest Point

At the highest point of the trajectory, the vertical component of the velocity is zero. The initial vertical velocity component is \( v_{y} = v \sin(\theta) \) which equals \( 25.0 \sin(53.13^\circ) \). To calculate this:\[v_{y} = 25.0 \times 0.8 = 20.0 \text{ m/s}.\]Using the equation \( v_{y} = u_{y} - gt \) at the highest point, \( v_{y} \) is zero. Thus, acceleration is \(-9.8 \text{ m/s}^2\) downward.

Step 4: Calculate the Water's Height Above the Ground When Hitting the Building

We can find the height using the vertical displacement formula \( y = v_{y}t - \frac{1}{2}gt^2 \):\[y = 20.0 \times 3.00 - \frac{1}{2} \times 9.8 \times (3.00)^2.\]Calculate:\[y = 60.0 - 44.1 = 15.9 \text{ m}.\]

Step 5: Determine the Speed of the Water Just Before It Hits the Building

To find the speed just before the water hits the building, use the Pythagorean theorem with final horizontal and vertical speeds. Horizontal speed \( v_{x} = 15.0 \text{ m/s} \) remains unchanged. Vertical speed just before impact, considering downward direction:\[v_{y} = u_{y} - gt = 20.0 - 9.8 \times 3.00 = -9.4 \text{ m/s}.\]Determine total speed: \[v = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(15.0)^2 + (-9.4)^2} \approx 17.7 \text{ m/s}.\]

Key Concepts

Angle of Elevation

The angle of elevation is a critical factor in projectile motion. It determines the initial direction at which an object is projected into the air.
For this problem involving water from the hose, the angle of elevation is the angle between the horizontal ground and the direction of the water's projection. Calculating the angle helps in targeting and maintaining a specific path and range.
To find the angle of elevation in this exercise, we used the horizontal motion formula set in terms of time (\[ x = v_x \cdot t \]), where the horizontal velocity component is expressed as \( v_x = v \cos(\theta) \).
Simplifying it for given values, the cosine of the angle of elevation came out as 0.6, which using an arccos calculator, converts to an approximate degree of 53.13°. This tells us the angle at which to aim the hose to reach the target distance.

Horizontal Motion

In projectile motion, horizontal motion refers to the movement along the horizontal axis. It occurs with constant velocity since there's no horizontal acceleration involved when air resistance is ignored.
The primary formula for horizontal motion is \( x = v_x \cdot t \), where \( x \) is the horizontal distance covered, \( v_x \) is the horizontal velocity, and \( t \) is the time.
For our water stream example, the horizontal velocity was calculated using \( v \cos(\theta) \). Given the conditions, the firemen needed to adjust the angle such that the water covers a horizontal distance of 45 m in 3 seconds.
This constant horizontal speed showcases how, despite moving upwards and downwards, the horizontal component remains unaffected by gravity.

Vertical Motion

Vertical motion, unlike horizontal motion, is affected by gravity. This influence of gravity causes an upward-moving object to slow down, stop, and eventually accelerate downwards.
The initial vertical velocity component here is given as \( v_y = v \sin(\theta) \), with 53.13° found as the angle.
At the highest point of its trajectory, the vertical velocity component temporarily reaches zero due to gravity's pull; this is where the water stops rising.
Using the equation \( y = v_y t - \frac{1}{2} gt^2 \), we found the vertical displacement at the building as 15.9 meters, showing the peak height the water reached before descending.

Velocity Components

Velocity components in projectile motion separate the initial velocity into two distinct parts – horizontal (\( v_x \)) and vertical (\( v_y \)).
Understanding this separation helps one determine projectile path properties since horizontal and vertical motions are independent.
In this scenario, the water's initial speed was 25 m/s. We found \( v_x = 25 \cos(53.13^\circ) \approx 15 \) meters per second. The vertical component is \( v_y = 25 \sin(53.13^\circ) \approx 20 \) meters per second.

• Horizontal Velocity: Remains constant throughout and means the water keeps moving forward at the same pace.
• Vertical Velocity: Changes over time due to gravity, reaching zero at the peak and altering when descending.

By combining these components, we manage trajectory and predict where and when the projectile hits the target.

Acceleration Due to Gravity

Gravity, commonly denoted as \( g \), significantly impacts vertical motion.
It accelerates objects downward at a rate of approximately \( 9.8 \text{ m/s}^2 \). This consistent acceleration affects the water's vertical velocity but not its horizontal velocity.
In the fire hose scenario, gravity caused a downward acceleration opposing the water's initial vertical velocity.

• At the top of the arc, the water's upward velocity stops, marking the influence of gravity maximally as it begins the descent.
• The final vertical velocity is calculated incorporating this gravitational acceleration over time.

On impact with the building, gravity has decelerated the water’s upward motion, then accelerated it downward, contributing to its overall speed on contact at a downward vertical speed of \(-9.4\) m/s, showing its decelerating impact.