A fire fighter is trying to shoot water straight to the window located at the second floor of a
house 6 m above the ground. The distance between the fire fighter and the house is 8 m and he
holds the fire hose 1.8 m above the ground. The water leaves the hose with a constant speed of
12.5 m/s. Initially, the fire fighter aims the hose at 53 ̊above the horizontal and misses the
window. (we can assume that the hose and the window are in the same vertical plane)
a. How much time it will take for the water flow to reach the house?
b. How far above the window does the water go?
c. What is the magnitude of the velocity of water when it strikes the house?
d. What must be the minimum angle and speed of the flow in order to get water
right into the window?
a. Well, it seems like that fire fighter needs some work on his aim. To find out how much time it will take for the water flow to reach the house, we can start by finding the vertical component of the water's velocity. Considering the initial angle of 53 degrees and a constant speed of 12.5 m/s, we can use some trigonometry to calculate that.
The vertical component of the velocity (Vy) can be found using the formula: Vy = V * sin(θ), where V is the initial speed and θ is the angle. So, Vy = 12.5 m/s * sin(53°).
Now that we have the vertical component of the velocity, we can use it to find out the time it takes for the water flow to reach the house. The vertical distance between the fire fighter and the house is 6 m, let's call it d.
We know that the formula to calculate the vertical distance in the presence of constant acceleration (which we can assume since gravity is acting on the water flow) is: d = Vi * t + (1/2) * a * t^2, where Vi is the initial velocity, t is the time, and a is the acceleration.
In this case, Vi is the vertical component of the velocity Vy, t is what we're trying to find, and a is the gravitational acceleration (approximately 9.8 m/s^2). The initial vertical velocity is Vy (which we calculated before), and the vertical distance d is 6 m. Plugging in the values, we get the following equation:
6 = Vy * t + (1/2) * (-9.8) * t^2.
To solve this quadratic equation, you can use the quadratic formula. In this case, it simplifies to a linear equation, and after solving it, you'll find the value of t.
b. To find out how far above the window the water goes, we need to find the vertical distance the water traveled after it reached the house. We already know the vertical distance of 6 m, but that's from the ground to the house. So, we need to subtract the height of the window, which is given as 2 m. Therefore, the water goes 6 m - 2 m above the window.
c. Now, let's figure out the magnitude of the velocity of the water when it strikes the house. We already know the horizontal component of the velocity (Vx) since it remains constant at 12.5 m/s, and the vertical component of the velocity (Vy) at that time.
The magnitude of the velocity can be found using the Pythagorean theorem: V = sqrt(Vx^2 + Vy^2).
Plug in the values of Vx and Vy that we've already calculated, and calculate the magnitude of the velocity V.
d. Finally, let's determine the minimum angle and speed of the flow in order to get the water right into the window. For this to happen, the water needs to reach the vertical distance of 6 m (from the ground to the house) and the horizontal distance of 8 m (from the fire fighter to the house).
We can use the range formula to find the minimum angle and speed. The range formula is: R = (V^2 * sin(2θ)) / g, where R is the horizontal distance, V is the speed of the flow, θ is the angle, and g is the gravitational acceleration (approximately 9.8 m/s^2).
In this case, we know the horizontal distance R (which is 8 m), the gravitational acceleration g, and we need to find the angle θ and the speed V.
Rearranging the formula, we get: V = sqrt((R * g) / sin(2θ)).
Plug in the values of R and g that we know, and calculate the minimum speed V. Then, to find the minimum angle θ, rearrange the formula again: θ = (1/2) * sin^(-1)((R * g) / V^2).
And there you have it, the minimum angle and speed required to get the water flow right into the window.
To solve this problem, we can break it down into several steps:
Step 1: Find the vertical distance between the fire hose and the window.
Given:
- Height of the house (h) = 6 m
- Height at which the fire fighter holds the hose (h_hose) = 1.8 m
Vertical distance = h - h_hose = 6 m - 1.8 m = 4.2 m
Step 2: Calculate the time it takes for the water flow to reach the house.
To find the time, we need to consider the vertical motion of the water.
Using the equation: h = ut + (1/2)gt^2
where:
- h = vertical distance
- u = initial vertical velocity (0, as the initial velocity is horizontal)
- g = acceleration due to gravity (-9.8 m/s^2)
- t = time
Substituting the values:
4.2 m = 0 t + (1/2)(-9.8 m/s^2) t^2
Rearranging the equation, we get:
4.9 t^2 = 4.2
t^2 = 4.2 / 4.9
t^2 ≈ 0.8571
t ≈ √(0.8571)
t ≈ 0.9264 s
So, it will take approximately 0.9264 seconds for the water flow to reach the house.
Step 3: Calculate how far above the window the water goes.
Since the water is projected at an angle above the horizontal, we need to find the vertical component of its initial velocity.
Using trigonometry:
Vertical component = velocity × sin(angle)
Vertical component = 12.5 m/s × sin(53°)
Vertical component ≈ 12.5 m/s × 0.7986
Vertical component ≈ 9.97 m/s
Now, we can determine the distance traveled vertically:
Vertical distance = (Vertical component) × (time)
Vertical distance ≈ 9.97 m/s × 0.9264 s
Vertical distance ≈ 9.2288 m
Therefore, the water will go approximately 9.2288 meters above the window.
Step 4: Find the magnitude of the velocity of the water when it strikes the house.
The magnitude of the velocity is the total velocity of the water, combining both the horizontal and vertical components.
Using Pythagoras' theorem:
Magnitude of velocity = √((horizontal component)^2 + (vertical component)^2)
Magnitude of velocity = √((12.5 m/s)^2 + (9.97 m/s)^2)
Magnitude of velocity ≈ √(156.25 m^2/s^2 + 99.4 m^2/s^2)
Magnitude of velocity ≈ √(255.65 m^2/s^2)
Magnitude of velocity ≈ 15.996 m/s
Therefore, the magnitude of the velocity of the water when it strikes the house is approximately 15.996 m/s.
Step 5: Calculate the minimum angle and speed of the flow to get water right into the window.
To get the water right into the window, the vertical distance traveled by the water should be equal to the height of the house.
Using the equation for vertical distance calculated in Step 3:
Vertical distance = 9.2288 m
Equating this to the height of the house, we have:
9.2288 m = 6 m
9.2288 m = (speed × sin(angle)) × (time)
9.2288 m = (speed × sin(angle)) × 0.9264 s
Simplifying the equation, we get:
speed × sin(angle) = 9.2288 m / 0.9264 s
speed × sin(angle) ≈ 9.965 m/s
Now, to minimize the angle, we want the sin(angle) to be maximum, which occurs when sin(angle) = 1.
So, we have:
speed = 9.965 m/s
Therefore, the minimum speed required to get the water right into the window is approximately 9.965 m/s. The angle is 90°, which means the water should be projected vertically upwards.
To solve this problem, we can apply the principles of projectile motion. Let's break it down step by step:
a. To find the time it takes for the water flow to reach the house, we can use the horizontal motion of the water. Since there is no acceleration in the horizontal direction, we can use the equation:
distance = speed * time
The horizontal distance between the firefighter and the house is 8 m. Since the firefighter is aiming 53 degrees above the horizontal, we need to find the horizontal component of the initial velocity. We can calculate this using the equation:
horizontal component of velocity = initial velocity * cos(angle)
where the angle is 53 degrees.
Plugging in the values, we have:
horizontal component of velocity = 12.5 m/s * cos(53 degrees)
Now we can use the formula for time:
time = distance / horizontal component of velocity
Plugging in the values, we have:
time = 8 m / (12.5 m/s * cos(53 degrees))
Calculate this to find the time it takes for the water to reach the house.
b. To find how far above the window the water goes, we need to consider the vertical motion of the water. We can use the equation:
vertical displacement = initial vertical velocity * time + (1/2) * acceleration * time^2
The initial vertical velocity can be calculated using:
initial vertical velocity = initial velocity * sin(angle)
where the angle is 53 degrees.
Since the water is projected upwards and the house is 6 m above the ground, the vertical displacement will be the sum of 6 m and the distance the water goes above the window. The acceleration in this case is due to gravity and is approximately 9.8 m/s^2.
Plugging in the values, we have:
vertical displacement = 6 m + initial vertical velocity * time + (1/2) * (-9.8 m/s^2) * time^2
Solve this equation to find how far above the window the water goes.
c. To find the magnitude of the velocity of the water when it strikes the house, we can use the equation for the magnitude of velocity in projectile motion:
velocity = square root of (horizontal component of velocity^2 + vertical component of velocity^2)
The horizontal component of velocity is the same as the horizontal component calculated in part (a). The vertical component of velocity can be calculated using:
vertical component of velocity = initial velocity * sin(angle)
where the angle is 53 degrees.
Plug in the values and calculate the magnitude of the velocity.
d. To find the minimum angle and speed of the flow in order to get the water right into the window, we need the vertical displacement to be equal to the height of the window, which is 6 m. Set up the equation:
vertical displacement = 6 m
Use the same equation as in part (b) and solve for the initial velocity and angle that satisfy this condition.
This will give you the minimum angle and speed required to get the water right into the window.