Let's break down the problem step-by-step: --- ## **Given Data** - Initial velocity when entering water, \( v_ = 16.5 \) ft/s - Acceleration in water: \( a = 10 - .8v \) (in \( \text{ft/s}^2 \), \( v \) in ft/s) - Time to reach bottom: \( t = 4.500 \) s - **Required:** Final speed \( v \) when the ball hits the bottom --- ## **Step 1: Write the Differential Equation** The acceleration is given as: \[ a = \frac{dv}{dt} = 10 - .8v \] --- ## **Step 2: Separate Variables and Integrate** \[ \frac{dv}{dt} = 10 - .8v \] \[ \Rightarrow \frac{dv}{10 - .8v} = dt \] Integrate both sides: \[ \int_{v_}^{v} \frac{dv}{10 - .8v} = \int_{}^{t} dt \] Let \( u = 10 - .8v \), so \( du = -.8 dv \), \( dv = \frac{du}{-.8} \): \[ \int_{v_}^{v} \frac{dv}{10 - .8v} = -\frac{1}{.8} \int_{u_}^{u} \frac{du}{u} \] \[ = -\frac{1}{.8} [\ln(u)]_{u_}^{u} \] \[ = -\frac{1}{.8} [\ln(10 - .8v) - \ln(10 - .8v_)] \] \[ = -\frac{1}{.8} \ln \left( \frac{10 - .8v}{10 - .8v_} \right) \] Set equal to \( t \): \[ -\frac{1}{.8} \ln \left( \frac{10 - .8v}{10 - .8v_} \right) = t \] --- ## **Step 3: Solve for Final Speed \( v \)** Let \( v_ = 16.5 \) ft/s, \( t = 4.5 \) s: \[ -\frac{1}{.8} \ln \left( \frac{10 - .8v}{10 - .8 \times 16.5} \right) = 4.5 \] \[ 10 - .8 \times 16.5 = 10 - 13.2 = -3.2 \] So: \[ -\frac{1}{.8} \ln \left( \frac{10 - .8v}{-3.2} \right) = 4.5 \] \[ \ln \left( \frac{10 - .8v}{-3.2} \right) = -.8 \times 4.5 \] \[ = -3.6 \] \[ \frac{10 - .8v}{-3.2} = e^{-3.6} \] \[ 10 - .8v = -3.2 e^{-3.6} \] \[ -.8v = -3.2 e^{-3.6} - 10 \] \[ .8v = 3.2 e^{-3.6} + 10 \] \[ v = \frac{3.2 e^{-3.6} + 10}{.8} \] --- ## **Step 4: Calculate the Value** Find \( e^{-3.6} \): \[ e^{-3.6} \approx .0273 \] \[ 3.2 \times .0273 \approx .08736 \] \[ v = \frac{.08736 + 10}{.8} \] \[ v = \frac{10.08736}{.8} \] \[ v \approx 12.609 \text{ ft/s} \] --- ## **Final Answer** \[ \boxed{12.6 \text{ ft/s}} \] --- ### **Summary Table** | Step | Description | Value/Formula | |------|------------------------------|------------------------| | 1 | Write acceleration equation | \( a = 10 - .8v \) | | 2 | Separate and integrate | See above | | 3 | Plug in values, solve for \(v\) | \( v \approx 12.6 \) ft/s | | 4 | **Final speed** | \(\boxed{12.6~\text{ft/s}}\) | --- **The speed of the ball when it hits the bottom of the lake is \(\boxed{12.6~\text{ft/s}}\).**

# Step-by-Step Solution ## Given Data - Initial velocity when entering water, \( v_0 = 16.5 \) ft/s - Acceleration in water: \( a = 10 - 0.8v \) (in ft/s², \( v \) in ft/s) - Time to reach bottom: \( t = 4.500 \) s - **Required:** Final speed \( v \) when the ball hits the bottom --- ## Step 1: Write the Differential Equation The acceleration is given as: \[ a = \frac{dv}{dt} = 10 - 0.8v \] --- ## Step 2: Separate Variables and Integrate Rearranging gives: \[ \frac{dv}{dt} = 10 - 0.8v \] \[ \Rightarrow \frac{dv}{10 - 0.8v} = dt \] Integrate both sides: \[ \int_{v_0}^{v} \frac{dv}{10 - 0.8v} = \int_{}^{t} dt \] Let \( u = 10 - 0.8v \), then \( du = -0.8 dv \), or \( dv = \frac{du}{-0.8} \): \[ \int_{v_0}^{v} \frac{dv}{10 - 0.8v} = -\frac{1}{0.8} \int_{u_0}^{u} \frac{du}{u} \] \[ = -\frac{1}{0.8} [\ln(u)]_{u_0}^{u} \] \[ = -\frac{1}{0.8} [\ln(10 - 0.8v) - \ln(10 - 0.8v_0)] \] \[ = -\frac{1}{0.8} \ln \left( \frac{10 - 0.8v}{10 - 0.8v_0} \right) \] Setting equal to \( t \): \[ -\frac{1}{0.8} \ln \left( \frac{10 - 0.8v}{10 - 0.8v_0} \right) = t \] --- ## Step 3: Solve for Final Speed \( v \) Substituting \( v_0 = 16.5 \) ft/s and \( t = 4.5 \) s: \[ -\frac{1}{0.8} \ln \left( \frac{10 - 0.8v}{10 - 0.8 \times 16.5} \right) = 4.5 \] Calculating: \[ 10 - 0.8 \times 16.5 = 10 - 13.2 = -3.2 \] Thus, \[ -\frac{1}{0.8} \ln \left( \frac{10 - 0.8v}{-3.2} \right) = 4.5 \] \[ \ln \left( \frac{10 - 0.8v}{-3.2} \right) = -0.8 \times 4.5 = -3.6 \] \[ \frac{10 - 0.8v}{-3.2} = e^{-3.6} \] \[ 10 - 0.8v = -3.2 e^{-3.6} \] \[ -0.8v = -3.2 e^{-3.6} - 10 \] \[ 0.8v = 3.2 e^{-3.6} + 10 \] \[ v = \frac{3.2 e^{-3.6} + 10}{0.8} \] --- ## Step 4: Calculate the Value 1. Calculate \( e^{-3.6} \): \[ e^{-3.6} \approx 0.0273 \] 2. Substitute into the equation: \[ 3.2 \cdot 0.0273 \approx 0.08736 \] \[ v = \frac{0.08736 + 10}{0.8} \] \[ v = \frac{10.08736}{0.8} \] \[ v \approx 12.609 \text{ ft/s} \] --- ## Final Answer The speed of the ball when it hits the bottom of the lake is: \[ \boxed{12.6 \text{ ft/s}} \] --- ### Summary Table | Step | Description | Value/Formula | |------|------------------------------|------------------------| | 1 | Write acceleration equation | \( a = 10 - 0.8v \) | | 2 | Separate and integrate | See above | | 3 | Plug in values, solve for \(v\) | \( v \approx 12.6 \) ft/s | | 4 | **Final speed** | \(\boxed{12.6~\text{ft/s}}\) |