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.

$$\begin{gathered} ma=mg-b_{3}v \\ 100\frac{dv}{dt}=75\left(9.81\right)-20v \\ \frac{dv}{dt}=\left(9.81\right)-0.2v \\ v\text{'}+0.2v=9.81 \\ v.e^{0.2t}=∫9.81e^{0.2t}dt \\ v.e^{0.2t}=49.05e^{0.2t}+ \end{gathered}$$

When the value of $v=0,  t=0,  then  c=-49.05$

$$\begin{gathered} v.e^{0.2t}=49.05e^{0.2t}+49.05 \\ v=-49.05e^{-0.2t}+49.05 \\ v=\left(1-e^{-0.2t}\right)49.05 \end{gathered}$$

When the value of $t=30,  then  v=48.93$.

$x_3\left(t\right)=\frac{mgt}{b_3}-\frac{m^{2}g}{{b^2}_3}\left(1-e^{\frac{-b_{3}t}{m}}\right)$

When $t=30,  then  x_3=1226.86m$

Therefore, when chute opens the parachutist is $3000-1226.86=1773.14 m$.

For finding the value of t apply:

$$\begin{gathered} ma=mg-b_{4}v \\ 75\frac{dv}{dt}=75\left(9.81\right)-100v \\ v\text{'}+1.33v=9.81                      ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

Solving equation (1) the value of $x\left(t\right)$.

$x\left(t\right)=9.81  t+39.12 \left(1-e^{-t}\right)$

When then $x=1773.14$then $t-176.76-3.98e^{-t}=0$.

$t=176.76$(Exponential part is too small)

Thus, the parachute opens after dropping from the helicopter$\mathbf{176}\mathbf{.}\mathbf{76}\mathbf{+}\mathbf{30}\mathbf{=}\mathbf{206}\mathbf{.}\mathbf{76}  \mathbf{sec}\mathbf{.}$

When the value of then $t=60\sec ,  v=49.05$then$x\left(60\right)=2697.75$

Therefore, the parachutist is $3000-2697.75=302.25 m$above so $x_4=302.25$.

Now, the value of $x \left(t\right)=9.81  t+39.12 \left(1-e^{-t}\right)$.

When $x_4=302.25$then $9.81  t-263.01-39.24 e^{-t}=0$and$t=206.7\sec .$

Hence, the parachute opens after 1 min dropping from the helicopter$26.81+60=88.81  \sec .$