math question?

Question

The velocity of a skydiver t seconds after jumping is given by
v(t)= [-(32/k)^1/2] x [(1-e^2t((32k)^1/2)) / (1+e^2t((32k)^1/2))]]
in meter/second, where k is a constant that depends upon the size and shape of the object and the density of the air. Find the therminal velocity with k=0.00064 and k=0.00128

Answer 1

The terminal velocity will be an asymptote of the function.
This can be found be letting t get very large.
When this happens the 1's in the second expression can be ignored and this expression becomes -1.
[(1-e^2t((32k)^1/2)) / (1+e^2t((32k)^1/2))] = [(-e^2t((32k)^1/2)) / (e^2t((32k)^1/2))]] = -1

So terminal V = [(32/k)^1/2]

for k=0.00064: Terminal V = 223.6 m/s
k=0.00128: Terminal V = 158.1 m/s