A fireman d = 33.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of θi = 21.0° above the horizontal as shown in the figure. If the speed of the stream as it leaves the hose is vi = 40.0 m/s, at what height will the stream of water strike the building?
horizontal problem:
u = 40 cos 21 for the whole time t
d = 33
so
t = 33/(40 cos 21)
vertical problem, same t
initial vertical velocity Vi = 40 sin 21
h = Vi t - 4.9 t^2
Thank you so much!!!!
Ah, firefighting physics, always making a splash! Let's see what we can do with this situation.
First, we need to break down the motion into horizontal and vertical components. The horizontal component is what will give us the distance, and the vertical component is what will give us the height.
The horizontal component of the velocity (vix) is given by the equation vix = vi * cos(θi). Plugging in the values, we get vix = 40.0 m/s * cos(21.0°).
The vertical component of the velocity (viy) is given by the equation viy = vi * sin(θi). Plugging in the values, we get viy = 40.0 m/s * sin(21.0°).
Now, let's focus on the horizontal component. The time it takes for the water to travel the distance is given by the equation t = d / vix. Plugging in the values, we get t = 33.0 m / vix.
Finally, let's focus on the vertical component. The height (h) the water will strike the building is given by the equation h = viy * t - (1/2) * g * t^2. Here, g represents the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values we found earlier, we get h = (40.0 m/s * sin(21.0°)) * t - (1/2) * (9.8 m/s^2) * t^2.
Now, all we need to do is substitute the value of t we found earlier into this equation to find the height where the stream of water will strike the building. Just be sure to bring an umbrella!
To find the height at which the stream of water will strike the building, we can use the projectile motion equations.
First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component of the velocity (Vx) remains constant throughout the motion, while the vertical component of the velocity (Vy) changes due to gravity.
Given that the speed of the stream as it leaves the hose is 40.0 m/s and the angle above the horizontal is 21.0°, we can find the initial vertical and horizontal velocities:
Vy = vi * sin(θi)
Vx = vi * cos(θi)
Substituting the given values:
Vy = 40.0 m/s * sin(21.0°)
Vx = 40.0 m/s * cos(21.0°)
Next, we need to find the time it takes for the stream of water to reach the building. The horizontal motion and vertical motion are independent of each other, so we can consider them separately.
For the vertical motion, we need to find the time it takes for the water stream to reach its peak before falling back downward. We can use the equation:
Vy = Voy - g * t
where Vy is the vertical component of the velocity, Voy is the initial vertical component of the velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since the water stream reaches its peak when its vertical velocity becomes zero, we can set Vy to zero:
0 = vi * sin(θi) - g * t
Using the values we already know:
0 = 40.0 m/s * sin(21.0°) - 9.8 m/s^2 * t
Now, solve for t:
t = (40.0 m/s * sin(21.0°)) / 9.8 m/s^2
Now that we have the time it takes for the water stream to reach its peak (which is the time it takes to reach the building), we can find the vertical distance traveled using the equation:
Δy = Voy * t - (1/2) * g * t^2
where Δy is the vertical distance traveled, Voy is the initial vertical component of the velocity, g is the acceleration due to gravity, and t is the time.
Substituting the values we already know:
Δy = 40.0 m/s * sin(21.0°) * [(40.0 m/s * sin(21.0°)) / 9.8 m/s^2] - (1/2) * 9.8 m/s^2 * [(40.0 m/s * sin(21.0°)) / 9.8 m/s^2]^2
Simplifying this equation will give you the answer.