Question

Hitting Bottom A helicopter hovers 500feet above a large open tank full of liquid (not water). A dense compact object weighing 160pounds is dropped (released from rest) from the helicopter into the liquid. Assume that air resistance is proportional to instantaneous velocityv while the object is in the air and that viscous damping is proportional to${\mathit{v}}^{\mathbf{2}}$ after the object has entered the liquid. For air take$\mathit{k}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{4}$, and for the liquid take$\mathit{k}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$.Assume that the positive direction is downward. If the tank is75 feet high, determine the time and the impact velocity when the object hits the bottom of the tank. [Hint: Think in terms of two distinct IVPs. If you use (13), be careful in removing the absolute value sign. You might compare the velocity when the object hits the liquid-the initial velocity for the second problem-with the terminal velocity${\mathit{v}}_{\mathbf{r}}$of the object falling through the liquid.]

Short answer

The time and the impact velocity when the object hits the bottom of the tank is t=6.82 seconds and$v_i=52.8ft/s$.

Step-by-step solution

Define the question

We have a dense object of a weight w=Ib is dropped from rest from a helicopter (500 feet high) in the air with a resistance is proportional to the velocity of the object $\left(k=\frac{1}{4}\right)$ into an open large tank (75 feet high) with a viscous damping is proportional to the square of the velocity (k=0.1), and we have to obtain the time needed for the object to reach the bottom of the large tank as the following technique.

Describe the cases.

In this problem, we have two cases : the first is being the object in the air and the second is being the object in the liquid
First, for the air we have the differential equation

$\begin{array}{rcl}m\frac{dv}{dt}& =& mg-kv\\ \frac{dv}{dt}& =& g-\frac{k}{m}v\end{array}$

Since we havelocalid="1663931350051" $g=32ft/s,k=0.25$andlocalid="1663938877624" $m=\frac{160}{32}=5Ib.s/ft$, then we have the D.E as

localid="1663931877424" $$\begin{gathered} \frac{dv}{dt}=32-\frac{0.25}{5}v \\ \frac{dv}{dt}=32-0.05v \end{gathered}$$

This differential equation is a first order and separable, then we can solve it as

$\begin{array}{rcl}\int \frac{1}{(32-0.05v)}dv& =& \int dt\\ -20\int \frac{(-0.05)}{(32-0.05v)}dv& =& \int dt\\ -20ln(32-0.05v)& =& t+c_1\\ e^{\ln (32-0.05v)}& =& e^{\frac{-1}{20}t+c_2}\\ 32-0.05v& =& c_3{e}^{-\frac{1}{20}t}\\ 0.05v& =& 32-c_3{e}^{-\frac{1}{20}t}\end{array}$

Then we have

$\begin{array}{rcl}v\left(t\right)& =& 640-c_3{e}^{-\frac{1}{20}t}\end{array}$

After that, to find the value of constant we have to apply the point of condition

(v, t) = (0, 0) into equation (1) as

$$\begin{gathered} 0=640-c{e}^0 \\ c=640 \end{gathered}$$

Step 3: Use the substitution and simplify

After that, substitute with the value of constant c into equation (1), then we have

$v\left(t\right)=640-640e^{-\frac{1}{20}t}$

is the velocity of the object during its drop into the air at time t.
Now, since we know that the height of the object has a relation between its velocity as

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

Then using equation we can obtain the height of the object at time t as

$\int ds=\int vdt$

Then

$\begin{array}{rcl}s\left(t\right)& =& \int \left(640-640e^{-\frac{1}{20}t}\right)dt\\ & =& 640\int dt-640\int e^{{}^{-\frac{1}{20}t}}dt\\ & =& 640t-640\times (-20)e^{{}^{-\frac{1}{20}t}}+h\\ & =& 640t+12800e^{{}^{-\frac{1}{20}t}}+h\end{array}$

Simplify the equation

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

$$\begin{gathered} 0=0+12800e^0+h \\ h=-12800 \end{gathered}$$

After that, substitute with the value of constant h into equation, then we have

role="math" $\begin{array}{rcl}s\left(t\right)& =& 640t+12800e^{-\frac{1}{20}t}-12800\end{array}$

is the height of the object above the large tank at time t.

Now, we have to obtain the time at which the object leaves the air and hits the liquid by substituting with into equation as

$\begin{array}{rcl}500& =& 640t+12800e^{-\frac{1}{20}t}-12800\\ 13300& =& 640t+12800e^{-\frac{1}{20}t}\end{array}$

Then by trial and error, we have

$t_1\approx 5.9seconds$

is the time needed for the object to hit the liquid.
Now we have to obtain the velocity of the object when it leaves the air and hits the liquid in the tank by substituting with t=5.9 ft/sec into equation as

$\begin{array}{rcl}v\left(t\right)& =& 640-640e^{-0.295}\\ & =& 163.5ft/s\end{array}$

is the velocity of the object when it hits the liquid (initial velocity of the second case).
Second, for the liquid we have the differential equation

$\begin{array}{rcl}m\frac{dv}{dt}& =& mg-k{v}^2\\ \frac{dv}{dt}& =& g-\frac{k}{m}{v}^2\end{array}$

Since we have g=32 ft/sec,

k=0.1 and

$m=\frac{160}{32}$

=5 lb.s/ft, then we have the D.E as

$$\begin{gathered} \frac{dv}{dt}=32-\frac{0.1}{5}{v}^2 \\ \frac{dv}{dt}=32-0.02v^2 \end{gathered}$$

Solve the equation using separation

This differential equation is a first order and separable, then we can solve it as

$\begin{array}{rcl}\int \frac{1}{32-0.02v}dv& =& \int dt\\ -50\int \frac{1}{(-1600+v^2)}dv& =& \int dt\\ \int \frac{1}{(v-40)(v+40)}dv& =& -\frac{1}{50}\int dt\\ \int \left(\frac{A}{(v-40)}+\frac{B}{v+40}\right)dt& =& -\frac{1}{50}t+c_1\\ \int \left(\frac{(40A+Av-40B+Bv)}{(v-40)(v+40)}\right)dv& =& -\frac{1}{50}t+c_1\\ \int \left(\frac{(40A-40B)+(A+B)v}{(v-40)(v+40)}\right)dv& =& -\frac{1}{50}t+c_1\\ \int \left(\frac{1}{80}\frac{1}{(v-40)}-\frac{1}{80}\frac{1}{(v+40)}\right)dv& =& -\frac{1}{50}t+c_1\end{array}$

$\begin{array}{rcl}\frac{1}{80}ln(v-40)-\frac{1}{80}ln(v+40)& =& -\frac{1}{50}t+c_1\\ ln\left(\frac{v-40}{v+40}\right)& =& -\frac{8}{5}t+c_2\\ e^{ln\left(\frac{v-40}{v+40}\right)}& =& e^{-\frac{8}{5}t+c_2}\\ \left(\frac{v-40}{v+40}\right)& =& c{e}^{-\frac{8}{5}t}\\ v-40& =& cv{e}^{{}^{-\frac{8}{5}t}}+40c{e}^{{}^{-\frac{8}{5}t}}\\ v-cv{e}^{{}^{-\frac{8}{5}t}}& =& 40c{e}^{{}^{-\frac{8}{5}t}}+40\\ v\left(1-c{e}^{{}^{-\frac{8}{5}t}}\right)& =& 40\left(1+c{e}^{{}^{-\frac{8}{5}t}}\right)\end{array}$

Then we have

$v\left(t\right)=\frac{40\left(1+c{e}^{-8/5t}\right)}{1-c{e}^{-8/5t}}$

After that, to find the value of constant we have to apply the point of condition into equation $(v,t)=(163.5,0)$as

$\begin{array}{rcl}163.5& =& 40\frac{(1+c{e}^0)}{(1-c{e}^0)}\\ 40+40c_3& =& 163.5-163.5c\\ 203.5c& =& 123.5\\ c& =& 0.6\end{array}$

After that, substitute with the value of constant c into equation (5), then we have

$v\left(t\right)=\frac{40\left(1+0.6e^{-{\displaystyle \frac{8}{5}}t}\right)}{1-0.6e^{-\frac{8}{5}t}}$

is the velocity of the object during its drop into the air at time t.
Now, since we know that the height of the object has a relation between its velocity as

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

Simplify for finding the height

Then using equation we can obtain the height of the object in the liquid at time t as

$\int ds=\int vdt$

Then

$$\begin{gathered} s\left(t\right)=40\int \left(\frac{1+0.6e^{-{\displaystyle \frac{8}{5}}t}}{1-0.6e^{-\frac{8}{5}t}}\right)dt \\ =40\int \left(\frac{1+0.6e^{-{\displaystyle \frac{8}{5}}t}-2+2}{1-0.6e^{-\frac{8}{5}t}}\right)dt \\ =-40\int \left(\frac{1-0.6e^{-{\displaystyle \frac{8}{5}}t}-2}{1-0.6e^{-\frac{8}{5}t}}\right)dt \\ =-40\int dt+80\int \left(\frac{1}{1-0.6e^{-\frac{8}{5}t}}\right)dt \\ =-40t+80\int \frac{e^{-\frac{8}{5}t}}{e^{-\frac{8}{5}t}-0.6}dt \end{gathered}$$

Now, if we set$u=e^{\frac{8}{5}t}-0.6$, then we have$du=\frac{8}{5}{e}^{\frac{8}{5}t}dt$, then we have

$\begin{array}{rcl}s\left(t\right)\&& =& -40t+80\int \frac{(5/8)}{u}du\\ & =& -40t+50\ln \left(u\right)+h\\ & =& -40t+50\ln (e^{\frac{8}{5}}t-0.6)+h)\end{array}$

After that, to find the value of constant h, we have to apply the point of conditioninto equation$(s,t)=(0,0)$as

$\begin{array}{rcl}0& =& -40t+50ln(e^0-0.6)+h\\ 0& =& -45.8+h\\ h& =& 45.8\end{array}$

After that, substitute with the value of constant h into equation ( 7), then we have

$s\left(t\right)=-40t+50ln\left(e^{\frac{8}{5}t}-0.6\right)+45.8$

is the height of the object in the large tank at time t.
Now, we have to obtain the time at which the object hits the bottom of the tank by substituting with s=75ft into equation (4) as

$$\begin{gathered} 75=-40t+50ln\left(e^{\frac{8}{5}t}-0.6\right)+45.8 \\ 29.2=-40t+50ln\left(e^{\frac{8}{5}t}-0.6\right) \end{gathered}$$

Then by trial and error, we have

$t_2\approx 0.92seconds$

is the time need to hit the ground from hitting the liquid,
Then from (a) and (b), we can obtain the time needed for the object to hit the ground as

$\begin{array}{rcl}t& =& t_1+t_2\\ & =& 5.9+0.92\\ & =& 6.82secondsrequired\end{array}$

Also, we have to obtain the impact velocity of the object when it hits the bottom of the tank by substituting with t=0.92 seconds into equation (6) as

$\begin{array}{rcl}{v}_i& =& 40\left(\frac{1+0.6e^{-1.472}}{1-0.6e^{-1.472}}\right)\\ & =& 40\left(\frac{1+0.1377}{1-1.377}\right)\\ & \approx & 52.8ft/srequired\end{array}$

is the velocity of the object when it hits the bottom of the tank.