Fire Hose Water Stream Physics Solution

Visual Analysis

The image displays a projectile motion scenario involving a fire hose. A nozzle at the origin $(0,0)$ emits a stream of water at an angle $\theta$, where the slope $m = \tan(\theta)$. The trajectory is a downward-opening parabola. The variables $x$ and $y$ represent the horizontal distance and vertical height of the water stream, respectively.

Answer

The horizontal range of the stream is $R = \frac{mv^2}{16(1+m^2)}$, and the maximum height for a fixed slope is $H_{max} = \frac{m^2v^2}{64(1+m^2)}$. To achieve the maximum distance, the firefighter should use a slope of $m=1$ ($\theta = 45^\circ$).

Explanation

1. Finding the Horizontal Range (Part a) The stream reaches the ground when the height $y = 0$. We solve the provided equation for $x$, excluding the trivial solution $x=0$. $0 = -16(1+m^2) \left(\frac{x}{v}\right)^2 + mx$ Dividing by $x$ (since $x \neq 0$): $\frac{16(1+m^2)x}{v^2} = m \implies x = \frac{mv^2}{16(1+m^2)}$ This formula calculates the total horizontal distance the water travels before hitting the ground. ⚠️ Note: This step is required on exams to define the range of a projectile.

2. Determining Maximum Height for Constant $m$ (Part b) The height $y$ is a quadratic function of $x$. The maximum height occurs at the vertex, which is halfway across the range, $x_{mid} = \frac{mv^2}{32(1+m^2)}$. Substitute $x_{mid}$ back into the original height equation: $y_{max} = -16(1+m^2) \left[\frac{mv^2}{32(1+m^2)v}\right]^2 + m\left[\frac{mv^2}{32(1+m^2)}\right]$ $y_{max} = \frac{m^2v^2}{64(1+m^2)}$ This represents the peak height the water reaches for a specific nozzle angle.

3. Optimizing Distance by Varying Slope (Part c) To find the slope $m$ that maximizes the range $x(m)$, we take the derivative of the range formula with respect to $m$ and set it to zero. $x(m) = \frac{v^2}{16} \cdot \frac{m}{1+m^2}$ Using the quotient rule: $\frac{dx}{dm} = \frac{v^2}{16} \left[ \frac{(1+m^2)(1) - m(2m)}{(1+m^2)^2} \right] = 0$ $1 - m^2 = 0 \implies m = 1$ Substituting $v = 80$: $x(1) = \frac{80^2}{16} \cdot \frac{1}{2} = 200$ feet. The graph is a curve starting at $(0,0)$, peaking at $m=1$, and approaching $0$ as $m \to \infty$.

4. Highest Point on a Building (Part d) We want to maximize $y$ with respect to $m$ for a fixed distance $x_0$. $\frac{dy}{dm} = -16\left(\frac{x_0}{v}\right)^2(2m) + x_0 = 0$ $32m\left(\frac{x_0}{v}\right)^2 = x_0 \implies m = \frac{v^2}{32x_0}$ Substitute this optimal $m$ back into the equation for $y$: $y_{peak} = -16(1+(\frac{v^2}{32x_0})^2)(\frac{x_0}{v})^2 + \frac{v^2}{32x_0}x_0 = \frac{v^2}{64} - \frac{16x_0^2}{v^2}$ This formula determines the maximum vertical reach at a specific horizontal distance.

5. Newton-Raphson Iteration (Part e) Known: $v=80, x_0=60, y=15$. We seek $m$ such that $f(m) = 0$. The equation: $15 = -16(1+m^2)\left(\frac{60}{80}\right)^2 + 60m \implies 15 = -9(1+m^2) + 60m$ $f(m) = -9m^2 + 60m - 24 = 0$ $f'(m) = -18m + 60$ Newton's formula: $m_{n+1} = m_n - \frac{f(m_n)}{f'(m_n)}$

   • $m_0 = 1$
   • $m_1 = 1 - \frac{27}{42} \approx 0.4444$
   • $m_2 = 1.6318$ (Error check: Newton-Raphson converges quickly for quadratics). Let's refine: $m_1 = 0.3571$ $m_2 = 0.4300$ $m_3 = 0.4334$ $m_4 = 0.4334$ The roots of $-9m^2 + 60m - 24 = 0$ are found via the quadratic formula: $m = \frac{-60 \pm \sqrt{3600 - 864}}{-18} \approx 0.4334$ and $6.2333$. Starting at $m=1$ leads to $m \approx 0.4334$.

Final Answer

The range of the water is $\frac{mv^2}{16(1+m^2)}$. The slope for maximum distance is $m=1$. For the specific conditions in part (e), the required slope is: $\boxed{m \approx 0.4334}$

Common Mistakes

• Dimensional Inconsistency: Forgetting that $v$ is in ft/s and $g=32$ ft/s² is already embedded in the constant $16$ ($1/2g = 16$).
• Vertex Confusion: Assuming maximum height occurs at the end of the range instead of the midpoint $x = R/2$.
• Algebraic Error: Forgetting to expand $(1+m^2)$ when differentiating the height in part (d).