A parachutist whose mass is 100 kgdrops from a helicopter hovering 3000 m above the ground and falls under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b₃=20 N-sec/mwhen the chute is closed andb₄=100 N-sec/m when the chute is open. If the chute does not open until 30 sec after the parachutist leaves the helicopter, after how many seconds will he hit the ground? If the chute does not open until 1 min after he leaves the helicopter, after how many seconds will he hit the ground?

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.

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

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$26.81+60=88.81 \sec .$