Question

Skydiving A skydiver weighs 125 pounds, and her parachute and equipment combined weigh another 35 pounds. After exiting from a plane at an altitude of 15,000 feet, she waits 15 seconds and opens her parachute. Assume that the constant of proportionality in the model in Problem 35 has the value k 5 0.5 during free fall and k 5 10 after the parachute is opened. Assume that her initial velocity on leaving the plane is zero. What is her velocity and how far has she traveled 20 seconds after leaving the plane? See Figure 3.1.14. How does her velocity at 20 seconds compare with her terminal velocity? How long does it take her to reach the ground? [Hint: Think in terms of two distinct IVPs.]

Short answer

$v\left(20\right)=16.0106\text{ft}/\text{s}$

Terminal velocity$v=16\text{ft}/\text{s}$

$s\left(20\right)=2510.29\text{ft}$

Time from opening the parachute to reach ground is $:t=13.1$minutes

Step-by-step solution

Definition of Series circuits

The linear differential equation for the current i(t),

$L\frac{di}{dt}+Ri=E\left(t\right)$

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

Differentiate the equation

We have a skydiver with a parachute weighs$w=(125+35)=160\text{Ib}$with a velocity$15,000\text{ft}$at a heightis exiting from a plane with an air resistance proportional to the velocity. After 15 seconds the parachute is opened, and this skydiver has the differential equation

$\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{mg}-\mathrm{kv}$

$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}$

with the initial condition

$v_0=0\text{ft}/\text{s}$

and we have to solve this differential equation as the following technique :

Separate the variables

This equation is a first order linear and separable differential equation, then we can separate variables and then do integration as

$\begin{array}{c}\frac{\mathrm{dv}}{\mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}}=\mathrm{dt}\\ \int \frac{1}{\mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}}\mathrm{dv}=\int \mathrm{d}\mathrm{t}\\ -\frac{\mathrm{m}}{\mathrm{k}}\int \frac{-\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{v}=\int \mathrm{d}\mathrm{t}\\ -\frac{\mathrm{m}}{\mathrm{k}}\ln \left(\mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}\right)=\mathrm{t}+{\mathrm{c}}_1\end{array}$

$\begin{array}{c}\ln \left(\mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}\right)=-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{t}+{\mathrm{c}}_2\\ {\mathrm{e}}^{\ln \left(\mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}\right)}={\mathrm{e}}^{\left(-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{t}+{\mathrm{c}}_2\right)}\\ \mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}={\mathrm{e}}^{{\mathrm{c}}_2}{\mathrm{e}}^{\frac{\mathrm{k}}{\mathrm{m}}\mathrm{t}}\\ \mathrm{g}-\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}={\mathrm{ce}}^{\frac{\mathrm{k}}{\mathrm{m}}\mathrm{t}}\end{array}$

$\frac{\mathrm{k}}{\mathrm{m}}\mathrm{v}=\mathrm{g}-{\mathrm{ce}}^{\frac{\mathrm{k}}{\mathrm{m}}\mathrm{t}}$

Then we have

$\mathrm{v}\left(\mathrm{t}\right)=\frac{\mathrm{mg}}{\mathrm{k}}-{\mathrm{se}}^{\frac{\mathrm{k}}{\mathrm{m}}\mathrm{t}}$

is the formula of velocity.

Now we have to know that we have two cases, the first is free fall without opening the parachute and the second with the parachute, then we can follow this technique:

First, for free fall: since we have$m=\frac{160}{32}=5,k=0.5$and $g=32\text{ft}/{\text{s}}^2$, then by substituting into equation (2), we can obtain

$v_f\left(t\right)=\frac{5\times 32}{0.5}-c{e}^{\frac{0.5}{5}t}$

$=320-c{e}^{-\frac{1}{10}t}$

Find the constant value

After that, to find the value of constant$s$, we have to apply the point of condition$(v,t)=(0,0)$into equation (3), then we have

$0=320-s{e}^0$

Then we have

$s=320$

After that, substitute with the value of constant$s$ into equation (2), then we have

$v\left(t\right)=320-320e^{\frac{1}{10}t}$

is the velocity of the skydiver with his equipment without opening parachute at time$t$.

Now, we have the distance measured from releasing point to the position of cannonball at timeis related to the velocity of the cannonball as

$\frac{ds}{dt}=v$

which is a linear and separable, then we can obtain the distance at time t as

$\int ds=\int vdt$

Then we have

$\begin{array}{c}{s}_f\left(t\right)=\int (320-320e^{-\frac{1}{10}t})dt\\ =320\int dt-320\int e^{-\frac{1}{10}t}dt\\ =320t-320(-10e^{-\frac{1}{10}t})+h\end{array}$

$=320t+3200\left(e^{\frac{1}{10}t}\right)+h$

Find the constant value h

After that, to find the value of constant$h$, we have to apply the point of condition$(s,t)=(0\text{ft},0$seconds) into equation (6), then we have

$0=320\left(0\right)+3200e^0+h$

Then we have

$h=-3200$

After that, substitute with the value of constant into equation (b), then we have $s\left(t\right)=320t+3200e^{-\frac{1}{10}t}-3200$is the distance for the skydiver exit point to its position without parachute at time $t$.

Second, now for falling with opening parachute : we have to find the value of the initial condition of velocity$\left(v_p\right)$and distance$S_p$for this case , (i.e$v\left(15\right)$and$s_p$), by substituting with$t=15$seconds into equations (4) and ( 7 ) respectively, then we have

$\begin{array}{c}v\left(15\right)=320-320e^{\frac{1}{10}\times 15}\\ =320-320e^{\frac{3}{2}}\end{array}$

$=248.6\text{ft}/\text{s}$

and

$\begin{array}{c}\mathrm{s}\left(15\right)=320\times 15+3200{\mathrm{e}}^{\frac{1}{10}\times 15}-3200\\ =4800+3200{\mathrm{e}}^{-\frac{3}{2}}-3200\end{array}$

$=2314\text{ft}$

Substitution and simplification

Now with since we have$k=10$, then we can have the velocity for skydiver with parachute using equation (2) as

$\begin{array}{rcl}{v}_p\left(t\right)& =& \frac{5\times 32}{10}-s_2{e}^{\frac{10}{5}t}\\ & =& 16-s_2{e}^{-2t}\end{array}$

After that, to find the value of constant $s_2$for the second case, we have to apply this point of condition $(v,t)=(248.6\text{ft}/\text{s},0$seconds ) into equation (8) as

$248.6=16-s_2{e}^0$

$s_2{e}^{-30}=16-248.6$

Then we have

$s_2=-232.6$

After that, substitute with the value of constant$s_2$into equation (8), then we have

$v_p\left(t\right)=16+232.6e^{-2t}$

is the velocity of skydiver where the parachute is opening at time$t$.

Now, to obtain the velocity after 20 seconds form exiting from the plane, substitute with$t=5$seconds into equation (9), then we have

$v=16+232.6e^{-10}$

$=16+0.0106$

$=16.0106\text{ft}/\text{s}$ (required )

and we can obtain the terminal velocity by substituting with$t=\infty$into equation (9) as

$\begin{array}{c}{v}_{\text{ter}}=16+232.6e^{-\infty }\\ =16+0\end{array}$

$=16\text{ft}/\text{srequired}$

which is less than the velocity at $t=5$seconds.

Also, we can obtain the distance in the second case with opening the parachute at time$t$using equation (9) as

$\begin{array}{c}{s}_p\left(t\right)=\int (16+232.6e^{-2t})dt\\ =16\int dt+232.6\int e^{2t}dt\end{array}$

$=16t+232.6\left(-\frac{1}{2}{e}^{-2t}\right)+h_2$

$=16t-116.3e^{-2t}+h_2$

After that, to find the value of constant$h_2$, we have to apply the point of condition$(s,t)=(2314\text{ft},0$seconds) into equation (6), then we have

$0=16\left(0\right)-116.3e^0+h_2$

Then we have

$h_2=116.3$

After that, substitute with the value of constant$h$into equation (b), then we have

$s_p\left(t\right)=16t-116.3e^{-2t}+116.3$

is the distance for the skydiver opening parachute point to its position at time$t$.

Now, to obtain the distance from after 20 seconds from leaving the plane, we have to follow this technique:

We have to obtain the distance from opening the parachute by substituting with $t=5$seconds into equation (11) as

$s\left(5\right)=16\times 5-116.3e^{10}+116.3$

$=196.29\text{ft}$

Now we can obtain the distance after 20 seconds as

$$\begin{gathered} \begin{array}{l}s\left(20\right)=s\left(5\right)+s\left(15\right)\\ =196.29+2314\end{array} \\ =2510.29\text{ftrequired} \end{gathered}$$

Finally, to obtain the time at which the skydiver reaches the ground from opening the parachute, we have to follow this technique :

We have to obtain the distance from opening the parachute to the ground as

$s_p=s\left(\text{height}\right)-s\left(15\right)$

role="math" localid="1667890476472" $=15000-2314$

role="math" localid="1667890498612" $=12686\text{ft}$

After that, we have to obtain the required time by substituting with$s=12686\text{ft}$into equation (11) as

$12686=16t-116.3e^{2t}+116.3$

$12569.7=16t-116.3e^{2t}$

Then by trial and error, we can obtain the required time as

$t\approx 785.5\text{seconds}\times \frac{1}{60}$

$\approx 13.1\text{minutesrequired}$

$v\left(20\right)=16.0106\text{ft}/\text{s}$

Terminal velocity$v=16\text{ft}/\text{s}$

$s\left(20\right)=2510.29\text{ft}$

Time from opening the parachute to reach ground is$:t=13.1$ minutes