Question

A firefighter, \(60 \mathrm{~m}\) away from a burning building, directs a stream of water from a ground-level fire hose at an angle of \(37^{\circ}\) above the horizontal. If the water leaves the hose at \(40.3 \mathrm{~m} / \mathrm{s}\), which floor of the building will the stream of water strike? Each floor is \(4 \mathrm{~m}\) high.

Short answer

Based on the given information, the stream of water will strike the 6th floor of the building.

Step-by-step solution

Identify known values

We know the following information: - The horizontal distance between the firefighter and the building \(d = 60\,\text{m}\) - The angle the water is being sprayed above the horizontal, \(\theta = 37^{\circ}\) - The speed the water leaves the hose, \(v_0 = 40.3\,\text{m/s}\) - The height of each floor, \(h_f = 4\,\text{m}\)

Find the horizontal and vertical components of the initial speed

We can use trigonometry to find the horizontal and vertical components of the initial speed. \(v_{0x} = v_0 \cos(\theta) = 40.3\,\text{m/s} \cos(37^{\circ}) \approx 32.2\,\text{m/s}\) \(v_{0y} = v_0 \sin(\theta) = 40.3\,\text{m/s} \sin(37^{\circ}) \approx 24.4\,\text{m/s}\)

Calculate the time it takes for the water to reach the building

We will use the horizontal component of the speed \(v_{0x}\) and the horizontal distance \(d\) to find the time it takes for the water to reach the building. \(d = v_{0x}t \Rightarrow t = \dfrac{d}{v_{0x}} = \dfrac{60\,\text{m}}{32.2\,\text{m/s}} \approx 1.86\,\text{s}\)

Calculate the vertical position of the water when it reaches the building

We will use the vertical component of the speed \(v_{0y}\), the time \(t\) we found in step 3, and the following equation of motion to determine the height the water reaches when it hits the building: \(h = v_{0y}t - \dfrac{1}{2}gt^2\) Where \(g = 9.81\,\text{m/s}^2\) is the acceleration due to gravity. \(h = (24.4\,\text{m/s})(1.86\,\text{s}) - \dfrac{1}{2}(9.81\,\text{m/s}^2)(1.86\,\text{s})^2 \approx 21.2\,\text{m}\)

Determine which floor the stream of water will strike

Using the height of each floor \(h_f\), divide the height \(h\) by \(h_f\) and round up to the nearest whole floor to find which floor the stream of water will strike. Number of Floors = \(\lceil \dfrac{h}{h_f} \rceil = \lceil \dfrac{21.2\,\text{m}}{4\,\text{m}} \rceil = 6\) The stream of water will strike the 6th floor of the building.

Key concepts

Trigonometry

Understanding projectile motion means mastering some basic trigonometry. Here, we use it to break down the initial speed of water into horizontal and vertical components.

• The horizontal component, denoted as \(v_{0x}\), involves using the cosine of the angle (in this case, \(37^{\circ}\)). This is because cosine relates an angle to the adjacent side over the hypotenuse, which gives us \(v_{0x} = v_0 \cos(37^{\circ})\).


• The vertical component, \(v_{0y}\), involves sine, since sine relates an angle to the opposite side of a triangle. Thus, \(v_{0y} = v_0 \sin(37^{\circ})\).

Trigonometry simplifies these complex motions by letting us deal with straightforward right triangles. This way, we can easily calculate how fast the water moves in any direction.

Physics Equations

Projectile motion problems often rely on key physics equations. One essential aspect is correctly applying these equations to figure out both distance and height.

• The equation \(d = v_{0x}t\) tells us how long it takes the water to travel 60 meters horizontally, leveraging constant horizontal speed.


• For vertical motion, we use \(h = v_{0y}t - \frac{1}{2}gt^2\). This combines the initial vertical velocity with the gravitational pull to calculate the water's vertical position as it flies.

Using these equations correctly is essential. Keep in mind that gravity only affects the vertical component, while the horizontal motion remains uniform unless obstructed.

Vertical and Horizontal Components

Projectile motion problems split into two simultaneous motions: vertical and horizontal. These two components are independent of each other yet happen concurrently.

• The horizontal component is straightforward, maintaining a constant speed assuming air resistance is negligible. It's calculated using the cosine of the launch angle.


• The vertical component is more dynamic, changing due to gravity. Initially, the water moves up but is gradually slowed down until it starts descending.

When tackling these components, consider them separately to predict projectile trajectory. Once resolved, they offer a combined scenic arc, characteristic of projectile motion, ensuring you know precisely where the water will land.