a sky diver jumps from a reasonable height above theground.The air resistance sh

Question

a sky diver jumps from a reasonable height above theground.The air resistance she experiences is porportional to hervelocity, and the constant of porpoortionality is 0.2.it can beshown the the downward velocity of the sky diver at time tis given by v(t)=80(1-e-0.2t) how do i find the terminal velocity algebraicly???? a sky diver jumps from a reasonable height above theground.The air resistance she experiences is porportional to hervelocity, and the constant of porpoortionality is 0.2.it can beshown the the downward velocity of the sky diver at time tis given by v(t)=80(1-e-0.2t) how do i find the terminal velocity algebraicly???? v(t)=80(1-e-0.2t) how do i find the terminal velocity algebraicly????

Explanation / Answer

Terminal Velocity in this case means the point at which thesky diver no longer accelerates due to gravity, as in, the rate atwhich they fall does not change because the force of gravity isbalanced with the force of air resistance. In simpler terms, thevelocity stops changing because there is no acceleration. Hope thishelps. The easiest way that I can imagine: v(t) = 80(1 - e^(-.2t) ) =80 - 80e^(-0.2t) we want that second term to get to a point where it doesn'treally change, so we will test different times to see theresults: v(1) =80 - 80e^(-0.2(1)) = 80 - 80e^(-0.2) v(5) =80 - 80e^(-0.2(5)) = 80 - 80e^(-1) = 80 -(80/e) v(50) = 80 - 80e^(-0.2(50)) = 80 - 80e^(-10) = 80 -(80/(e^10)) as you can see, as time goes up, the second term keeps gettingsmaller. We can also take the limit as time goes to infinity: lim(t->) 80 - 80e^(-0.2t) =lim(t->) 80 - 80(1/(e^0.2t)) = 80 - 80(1/(e^)) = 80 - 80(1/) = 80 - 0 =80 using the limit we can determine that the terminal velocity(as time goes to infinity) is 80 (insert your units here)

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