A parachutist whose mass is 75 kg drops from a helicopter hovering 2000 m above the ground and falls toward the ground 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 b₁ = 30 N-sec/m when the chute is closed and b₂= 90 N-sec/m when the chute is open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will she reach the ground?

${\text{ma = mg -b}}_{\text{1}}\text{v}$

$$\begin{gathered} 75\frac{\text{dv}}{\text{dt}}\mathbf{=}75\left(9.81\right)-\left(30\text{v}\right) \\ \frac{\mathrm{dv}}{\mathrm{dt}}=(9.81)-0.4v \\ \mathrm{v}\text{'}+0.4\mathrm{v}=9.81 \\ \mathrm{v}.{\mathrm{e}}^{0.4\mathrm{t}}=\int 9.81{\mathrm{e}}^{0.4\mathrm{t}}\mathrm{dt} \\ \mathrm{v}.{\mathrm{e}}^{0.4\mathrm{t}}=24.52{\mathrm{e}}^{0.4\mathrm{t}}+\mathrm{c} \end{gathered}$$

When v=0 and t=0 , c=-24.52

$$\begin{gathered} \mathrm{v}.{\mathrm{e}}^{0.4\mathrm{t}}=24.52{\mathrm{e}}^{0.4\mathrm{t}}+24.52 \\ \mathrm{v}=-24.52{\mathrm{e}}^{-0.4\mathrm{t}}+24.52 \\ \mathrm{v}=(1-{\mathrm{e}}^{-0.4\mathrm{t}})24.52 \\ \mathrm{x}\left(\mathrm{t}\right)=24.52\mathrm{t}+61.3{\mathrm{e}}^{-0.4\mathrm{t}}+\mathrm{c} \end{gathered}$$

When x=20 and t=0 ,c=41.3

$\mathrm{x}\left(\mathrm{t}\right)=24.52\mathrm{t}+61.3{\mathrm{e}}^{-0.4\mathrm{t}}+41.3$

put x=0,

$$\begin{gathered} 0=24.52\mathrm{t}+61.3{\mathrm{e}}^{-0.4\mathrm{t}}+41.3 \\ 41.3=24.52\mathrm{t}+61.3{\mathrm{e}}^{-0.4\mathrm{t}} \end{gathered}$$

${\text{ma = mg -b}}_2\text{v}$

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}75\frac{\mathrm{dv}}{\mathrm{dt}}=75(9.81)-90\mathrm{v} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=(9.81)-1.2\mathrm{v} \\ \mathrm{v}\text{'}+1.2\mathrm{v}=9.81 \\ \mathrm{v}.{\mathrm{e}}^{1.2\mathrm{t}}=\int 9.81{\mathrm{e}}^{1.2\mathrm{t}}\mathrm{dt} \\ \mathrm{v}.{\mathrm{e}}^{1.2\mathrm{t}}=8.175{\mathrm{e}}^{1.2\mathrm{t}}+\mathrm{c} \end{gathered}$$

When v=0 and t=0 , c=-8.175

$$\begin{gathered} \mathrm{v}.{\mathrm{e}}^{01.2\mathrm{t}}=8.175{\mathrm{e}}^{1.2\mathrm{t}}+8.175 \\ \mathrm{v}=-8.175{\mathrm{e}}^{-1.2\mathrm{t}}+8.175 \\ \mathrm{v}=(1-{\mathrm{e}}^{-1.2\mathrm{t}})8.175 \\ \mathrm{x}\left(\mathrm{t}\right)=8.175\mathrm{t}+6.8125{\mathrm{e}}^{-0.4\mathrm{t}}+\mathrm{c} \end{gathered}$$

When x=20 and t=0, c=13.1875

$\mathrm{x}\left(\mathrm{t}\right)=8.175\mathrm{t}+6.8125{\mathrm{e}}^{-1.2\mathrm{t}}+13.1875$

put x=0,

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}0=8.175\mathrm{t}+6.8125{\mathrm{e}}^{-1.2\mathrm{t}}+13.1875 \\ 13.1875=8.175\mathrm{t}+6.8125{\mathrm{e}}^{-0.4\mathrm{t}} \end{gathered}$$

By trial and error method t = 236.885 sec.

The total time takes by the parachutist is 4.225 + 236.885 = 241.1 sec

Hence parachutist will reach the ground in 241 sec.