Let's solve the problem step by step: --- ## **Step 1: Given Information** - Initial speed of the ball as it enters the water: \( v_ = 16.5 \) ft/s - Acceleration in water: \( a = 10 - .8v \), where \( a \) in ft/s², \( v \) in ft/s - Time to reach the bottom: \( t = 4.500 \) s - We need to find the **depth \( d \) of the lake**. --- ## **Step 2: Set Up the Differential Equation** The acceleration given is: \[ a = \frac{dv}{dt} = 10 - .8v \] --- ## **Step 3: Solve the Differential Equation** Separate variables: \[ \frac{dv}{10 - .8v} = dt \] Integrate both sides: \[ \int \frac{dv}{10 - .8v} = \int dt \] Let \( u = 10 - .8v \Rightarrow du = -.8 dv \Rightarrow dv = -\frac{1}{.8} du \): \[ \int \frac{dv}{10 - .8v} = -\frac{1}{.8} \int \frac{du}{u} = -1.25 \ln|u| = -1.25 \ln|10 - .8v| \] So: \[ -1.25 \ln|10 - .8v| = t + C \] Solve for \( v \): \[ \ln|10 - .8v| = -.8t + C' \] \[ 10 - .8v = A e^{-.8t} \] --- ## **Step 4: Initial Condition** At \( t = \), \( v = 16.5 \) ft/s: \[ 10 - .8(16.5) = A e^{} \] \[ 10 - 13.2 = A \] \[ A = -3.2 \] So, \[ 10 - .8v = -3.2 e^{-.8t} \] \[ .8v = 10 + 3.2 e^{-.8t} \] \[ v = \frac{10 + 3.2 e^{-.8t}}{.8} \] \[ v = 12.5 + 4 e^{-.8t} \] --- ## **Step 5: Find Displacement (Depth)** \[ v = \frac{dx}{dt} = 12.5 + 4 e^{-.8t} \] Integrate to find \( x \) (from \( t = \) to \( t = 4.5 \)): \[ dx = \left(12.5 + 4 e^{-.8t}\right) dt \] \[ x = \int_^{4.5} \left(12.5 + 4 e^{-.8t}\right) dt \] Integrate: \[ x = \int_^{4.5} 12.5\,dt + \int_^{4.5} 4 e^{-.8t}\,dt \] First term: \[ \int_^{4.5} 12.5\,dt = 12.5 \times 4.5 = 56.25 \] Second term: \[ \int_^{4.5} 4 e^{-.8t}\,dt = 4 \times \left[ \frac{e^{-.8t}}{-.8} \right]_^{4.5} \] \[ = 4 \times \left( -1.25 \right) \left( e^{-.8 \times 4.5} - e^{} \right) \] \[ = -5 \left( e^{-3.6} - 1 \right) \] Calculate \( e^{-3.6} \): \[ e^{-3.6} \approx .02732 \] So: \[ -5 \left( .02732 - 1 \right) = -5 \times (-.97268) = 4.8634 \] Add both terms: \[ x = 56.25 + 4.8634 = 61.1134\ \text{ft} \] --- ## **Final Answer** \[ \boxed{61.1\ \text{ft}} \] **The depth of the lake is approximately 61.1 ft.**

# Step-by-Step Solution to Determine the Depth of the Lake ## Step 1: Given Information - **Initial speed of the ball entering the water:** \( v_0 = 16.5 \) ft/s - **Acceleration equation in water:** \( a = 10 - 0.8v \) (where \( a \) is in ft/s² and \( v \) is in ft/s) - **Time to reach the bottom:** \( t = 4.5 \) s - **Objective:** Find the **depth \( d \)** of the lake. ## Step 2: Set Up the Differential Equation The acceleration can be expressed as: \[ a = \frac{dv}{dt} = 10 - 0.8v \] ## Step 3: Solve the Differential Equation ### Separate Variables \[ \frac{dv}{10 - 0.8v} = dt \] ### Integrate Both Sides Using the substitution \( u = 10 - 0.8v \), then \( du = -0.8 dv \): \[ \int \frac{dv}{10 - 0.8v} = -\frac{1}{0.8} \int \frac{du}{u} \] This simplifies to: \[ -1.25 \ln|10 - 0.8v| = t + C \] ### Solve for \( v \) Rearranging gives: \[ \ln|10 - 0.8v| = -0.8t + C' \] Exponentiating both sides: \[ 10 - 0.8v = A e^{-0.8t} \] ## Step 4: Apply the Initial Condition At \( t = 0 \), \( v = 16.5 \) ft/s: \[ 10 - 0.8(16.5) = A \] Calculating: \[ 10 - 13.2 = A \implies A = -3.2 \] ### Substitute Back \[ 10 - 0.8v = -3.2 e^{-0.8t} \] \[ 0.8v = 10 + 3.2 e^{-0.8t} \implies v = \frac{10 + 3.2 e^{-0.8t}}{0.8} \] \[ v = 12.5 + 4 e^{-0.8t} \] ## Step 5: Find Displacement (Depth) ### Relate Velocity to Displacement \[ v = \frac{dx}{dt} = 12.5 + 4 e^{-0.8t} \] Integrating \( dx \): \[ dx = \left(12.5 + 4 e^{-0.8t}\right) dt \] Integrate from \( t = 0 \) to \( t = 4.5 \): \[ x = \int_0^{4.5} \left(12.5 + 4 e^{-0.8t}\right) dt \] ### Calculate the Integrals 1. **First integral**: \[ \int_0^{4.5} 12.5\,dt = 12.5 \times 4.5 = 56.25 \] 2. **Second integral**: \[ \int_0^{4.5} 4 e^{-0.8t}\,dt = 4 \left[-\frac{1}{0.8} e^{-0.8t}\right]_0^{4.5} = -5 \left( e^{-3.6} - 1 \right) \] Calculate \( e^{-3.6} \): \[ e^{-3.6} \approx 0.02732 \] So: \[ -5 \left( 0.02732 - 1 \right) = -5 \times (-0.97268) = 4.8634 \] ### Total Displacement Combine both results: \[ x = 56.25 + 4.8634 = 61.1134\ \text{ft} \] ## Final Answer The depth of the lake is approximately: \[ \boxed{61.1\ \text{ft}} \]