91Ó°ÊÓ

Use Newton’s method to approximate the time t when the object strikes the ground

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

The given values are $m=500$, $v_0=0$, $g=9.81$, $v_0=0$, $b=50$,

The equation of motion is $x\left(t\right)=\frac{mgt}{b}+\frac{m}{b}\left(v-\frac{mg}{b}\right)\left(1-e^{\frac{-bt}{m}}\right)         ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Put all the given values in (1)

$\frac{500\left(9.81\right)t}{50}+\frac{500}{50}\left(0-\frac{500\left(9.81\right)}{50}\right)\left(1-e^{\frac{-50t}{500}}\right)$

role="math" localid="1664162976240" $\mathbf{x} \left(\mathbf{t}\right)\mathbf{=}\mathbf{98}\mathbf{.}\mathbf{1}  \mathbf{t}\mathbf{+}\mathbf{981}  {\mathbf{e}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{10}}}\mathbf{-}\mathbf{981}  \mathbf{m}$

Put the value of $x\left(t\right)=1000 m$then

$$\begin{gathered} 1000=98.1t+981e^{\frac{-t}{10}}-981m \\ 1000+981=98.1t+981e^{\frac{-t}{10}} \\ 1981=98.1t+981e^{\frac{{\displaystyle -t}}{{\displaystyle 10}}} \\ t=\frac{1981}{98.1} \\ t=20.2 \end{gathered}$$

(Ignoring the exponential term is too large)

Therefore, the time $\mathbf{t}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{2} \mathbf{Sec}\mathbf{.}$

Let$f\left(t\right)=1981=98.1t+981e^{\frac{-t}{10}}=0$.

The Newton’s method is$t_{n+1}=t_n-\frac{f\left(t_n\right)}{f\text{'}\left(t_n\right)}$.

$f\left(t\right)=98.1-98.1e^{\frac{-t}{10}}=\left(1-e^{\frac{-t}{10}}\right)98.1$

And the formula is$t_{n+1}=t_n-\frac{f\left(t_n\right)}{f\text{'}\left(t_n\right)}$.

Put the value is$n=0$,$t_1=\frac{20.1936-\left(20.1936+10e^{\left(-2.011936\right)}-20.1936\right)}{\left(1-e^{-2.011936}\right)}$.

$\left(t_0=\frac{1981}{98.1}=20.1936\right)$

Therefore, the result is role="math" localid="1664163445467" ${\mathbf{t}}_{\mathbf{1}}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{663121} \mathbf{Sec}\mathbf{.}$