QuestionSuppose 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. $$v(t)=55-55 e^{-0.26 t}$$
How many seconds after diving will the sky diver's velocity be 41 meters per second? Round your answer to the nearest tenth, and do not round any intermediate computations. $\square$ seconds