Let's break down the problem and solve it step by step. --- ## **Problem Restatement** A falling object from rest (no wind, only air resistance proportional to the speed) has its velocity described by: \[ v(t) = \sqrt{\frac{mg}{k}} \tanh \left( \sqrt{\frac{kg}{m}} t \right) \] Where: - \( m \) = mass in kg - \( g = 9.8\, \text{m/s}^2 \) (acceleration due to gravity) - \( k \) = physical constant - Person has \( m = 80\, \text{kg} \) and \( k = .25\, \text{kg/s} \) - How long to reach \( v = 50\, \text{m/s} \)? --- ## **Step 1: Plug in the Known Values** Given: - \( m = 80 \) - \( g = 9.8 \) - \( k = .25 \) - \( v(t) = 50 \) The equation becomes: \[ 50 = \sqrt{\frac{80 \times 9.8}{.25}} \tanh \left( \sqrt{\frac{.25 \times 9.8}{80}} t \right) \] --- ## **Step 2: Simplify the Square Roots** ### Calculate \(\sqrt{\frac{mg}{k}}\): \[ \frac{mg}{k} = \frac{80 \times 9.8}{.25} = \frac{784}{.25} = 3136 \] \[ \sqrt{3136} = 56 \] ### Calculate \(\sqrt{\frac{kg}{m}}\): \[ \frac{kg}{m} = \frac{.25 \times 9.8}{80} = \frac{2.45}{80} = .030625 \] \[ \sqrt{.030625} = .175 \] So now the equation is: \[ 50 = 56 \tanh(.175 t) \] --- ## **Step 3: Solve for \( t \)** \[ \frac{50}{56} = \tanh(.175 t) \] \[ .893 = \tanh(.175 t) \] Take the inverse hyperbolic tangent (\(\tanh^{-1}\)) of both sides: \[ .175 t = \tanh^{-1}(.893) \] Recall: \[ \tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right) \] So, \[ \tanh^{-1}(.893) = \frac{1}{2} \ln\left(\frac{1+.893}{1-.893}\right) = \frac{1}{2} \ln\left(\frac{1.893}{.107}\right) \] \[ \frac{1.893}{.107} \approx 17.7 \] \[ \ln(17.7) \approx 2.87 \] \[ \frac{1}{2} \times 2.87 \approx 1.435 \] So, \[ .175 t \approx 1.435 \] \[ t \approx \frac{1.435}{.175} \approx 8.2 \text{ seconds} \] --- ## **Final Answer** \[ \boxed{t \approx 8.2\ \text{seconds}} \] **It takes about 8.2 seconds for the person to reach 50 m/s.**