91Ó°ÊÓ

Use Newton’s method to solve for t.

$t_{n+1}=t_n-\frac{f\left(t_n\right)}{f\text{'}\left(t_n\right)}$

For finding the weight of the object apply:

Net force=W-drag force

$$\begin{gathered} ma=W-16v \\ ma=mg-16v \\ 8\frac{dv}{dt}=8(9.81)-16v \\ 8v\text{'}=78.4-16v \end{gathered}$$

Further solve the above expression,

role="math" localid="1663943376324" $$\begin{gathered} v\text{'}=9.8-2v         \left(\text{Integrating factor} e^{2t}\right) \\ v\text{'}+2v=9.8 \\ v.e^{2t}=∫9.8e^{2t}dt \\ v.e^{2t}=4.9e^{2t}+C \\ v=4.9e^{2t}+C{e}^{-2t} \\ v.e^{2t}=4.9e^{2t}+C \end{gathered}$$

At the value of $v=-20  and  t=0,  then  C=24.9$ (c is constant and arrange negative sign)

$v=4.9+24.9e^{-2t}$

$$\begin{gathered} v=\frac{dx}{dt} \\ \frac{dx}{dt}=4.9+24.9e^{-2t} \\ x\left(t\right)=4.9t-12.45e^{-2t}+c \end{gathered}$$

Put the value of $t=0,  x=100,  then  C=112.45$

$x\left(t\right)=4.9 t-12.45 e^{-2t}+112.45$

When the value of $x=0$

$0=4.9t-12.45e^{-2t}+112.45$

By trial and error, the value of t $t=22.9\sec .$

Therefore, the value of t is $t=22.9\sec .$