Step 1 :Suppose that the velocity \(v(t)\) (in meters per second) of a sky diver falling near the Earth's surface is given by the following exponential function, where time \(t\) is the time after diving measured in seconds: \(v(t)=55-55 e^{-0.24 t}\).
Step 2 :We are asked to find how many seconds after diving will the sky diver's velocity be 41 meters per second.
Step 3 :We can solve this by setting \(v(t)\) equal to 41 and solving for \(t\).
Step 4 :Doing so gives us the equation \(41 = 55 - 55 e^{-0.24 t}\).
Step 5 :We can solve this equation for \(t\) using the properties of logarithms.
Step 6 :First, we subtract 41 from both sides to get \(0 = 14 - 55 e^{-0.24 t}\).
Step 7 :Then, we divide both sides by -55 to get \(-\frac{14}{55} = e^{-0.24 t}\).
Step 8 :Taking the natural logarithm of both sides gives us \(\ln(-\frac{14}{55}) = -0.24 t\).
Step 9 :Finally, we divide both sides by -0.24 to solve for \(t\).
Step 10 :Doing so gives us \(t = \frac{\ln(-\frac{14}{55})}{-0.24}\).
Step 11 :Rounding to the nearest tenth, we find that \(t \approx 5.7\).
Step 12 :Final Answer: The sky diver's velocity will be 41 meters per second approximately \(\boxed{5.7}\) seconds after diving.