Determining terminal velocity using exponential functions I am having trouble with this problem.  Any help would be appreciated.
The velocity v(t) of a $184$ pound skydiver is approximated by
$v(t)=\frac{184}{1.1}(1−e^{−24.2t/184})$
where $t$ is in seconds after jumping from a plane.
What is the skydiver's terminal velocity? That is, in terms of the constants of the problem $\lim_{t\to\infty}v(t)= $?
Also, How long after jumping does it take the skydiver to reach $72$ percent of the terminal velocity? 
I really do not know where to start with this problem.  Thanks in advance.
 A: Hint:
The equation of  velocity has the form:
$$
v(t)=a \left(1-e^{-bt}\right) \qquad with : \quad a,b >0
$$
As noted in the comment the final velocity is:
$$
v_F=\lim_{t \to \infty}a \left(1-e^{-bt}\right)=a
$$
and $v$ is $72\%$ of $v_F$, i.e. $v=\dfrac{72}{100}a$ if:
$$
\dfrac{72}{100}a=a \left(1-e^{-bt}\right)
$$
i.e.
$$
\dfrac{72}{100}=1-e^{-bt} \qquad \iff \qquad e^{-bt}=\dfrac{28}{100}
$$
can you solve this equation in $t$  ?
A: Notice, the terminal velocity of the skydiver is given as 
$$\lim_{t\to \infty}v(t)$$$$=\lim_{t\to \infty}\frac{184}{1.1}(1-e^{-24.2t/184})$$$$=\frac{184}{1.1}\lim_{t\to \infty}(1-e^{-24.2t/184})$$$$=\frac{184}{1.1}(1-0)=\color{blue}{\frac{1840}{11}}$$
Now, the velocity $v(t)$ becomes $72$% of the terminal velocity hence we have 
$$\frac{72}{100}\frac{1840}{11}=\frac{184}{1.1}(1-e^{-24.2t/184})$$
$$e^{-24.2t/184}=1-0.72=0.28=\frac{7}{25}$$
$$\frac{24.2t}{184}=\ln\left({\frac{25}{7}}\right)$$
$$t=\frac{184}{24.2}\ln\left({\frac{25}{7}}\right)=\color{blue}{9.68\ sec}$$
