SOLUTION: The velocity v of an object t seconds after it has been dropped from a height above the surface of the earth is given by the equation v=32t feet per second, assuming no air resista

Question 623704: The velocity v of an object t seconds after it has been dropped from a height above the surface of the earth is given by the equation v=32t feet per second, assuming no air resistance. If we assume that air resistance is proportional to the square of the velocity, then the velocity after t seconds is given by the following equation.
v=100((e^.64t-1)/(e^.64t+1))
In how many seconds will the velocity be 50 feet per second?
Getting the answer would be nice, but I would really appreciate an explanation as well. Thank you !
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
v=100((e^.64t-1)/(e^.64t+1))
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100((e^.64t-1)/(e^.64t+1)) = 50
((e^.64t-1)/(e^.64t+1)) = 0.5
e^.64t-1 = 0.5*(e^.64t + 1) = 0.5e^64t + 0.5
e^.64t - 0.5*e^.64t = 1.5
0.5*e^.64t = 1.5
e^.64t = 3
0.64t = ln(3)
t = ln(3)/0.64
t =~ 1.7166 seconds
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