Question

Repeat Problem 36, but this time assume that air resistance is proportional to instantaneous velocity. It stands to reason that the maximum height attained by the cannonball must be less than that in part (b) of Problem 36. Show this by supposing that the constant of proportionality is $\mathit{k}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0025}$. [Hint: Slightly modify the differential equation in Problem 35.]

Short answer

$v\left(t\right)=6700e^{-\frac{1}{2000}t}-6400$

$s\left(t\right)=-6400t-1340000e^{\frac{1}{200}t}+1340000$

Maximum height $=1363.8\mathrm{ft}$

Step-by-step solution

Definition of Series circuits

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

$\mathit{L}\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{R}\mathit{i}\mathbf{=}\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

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 small cannonball with a weight $w=16$Ib with a velocityis shotted vertically upward with an air resistance proportional to the velocity of this cannonball, has the differential equation

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

with the initial condition

$v_0=300\text{ft}/\text{s}$

and we have to show that the maximum height for the cannonball must be less than that in problem (36) as the following technique :

Since we have $g=32\text{ft}/{\text{s}}^2,m=\frac{w}{g}=\frac{16}{32}=0.5$and $k=0.0025,$then we can have the differential equation shown in (1) as

$\begin{array}{rcl}0.5\frac{dv}{dt}& =& -0.5\times 32-0.0025v\times 2\\ \frac{dv}{dt}& =& -32-0.005v\end{array}$

Separate the variables and integrate

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

$\frac{dv}{-32-0.005v}=dt$

$\begin{array}{rcl}-\int \frac{1}{32+0.005v}dv& =& \int dt\\ -200\int \frac{0.005}{32+0.005v}dv& =& \int dt\\ -200\ln (32+0.005v)& =& t+c_1\\ \ln (32+0.005v)& =& -\frac{1}{200}t+c_2\\ e^{\ln (32+0.005v)}& =& e^{\left(-\frac{1}{200}t+c_2\right)}\\ 32+0.005v& =& e^{c_2}{e}^{\frac{1}{200}t}\\ 0.005v& =& c{e}^{\frac{1}{200}t}-32\end{array}$

After that, to find the value of constant k, we have to apply the point of condition$(v,t)=(300,0)$into equation (2), then we have

$300=k{e}^0-6400$

Then we have

$\begin{array}{rcl}k& =& 300+6400\\ & =& 6700\end{array}$

After that, substitute with the value of constantinto equation (2), then we have

$v\left(t\right)=6700e^{\frac{1}{2200}t}-6400$

is the velocity of the small cannonball at time t.

Find the distance

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

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

with the condition

$s(t=0)=0$

This differential equation (a) is a linear and separable, then we can obtain the distance at time t as

$\int ds=\int vdt$

Then we have

$\begin{array}{rcl}s& =& \int \left(6700e^{\frac{1}{2000}t}-6400\right)dt\\ & =& \int 6700e^{-\frac{1}{2000}t}dt-\int 6400dt\\ & =& 6700\int e^{-\frac{1}{200}t}dt-6400\int dt\\ & =& 6700\left(-200e^{\frac{1}{200}t}\right)-6400t+h\\ & =& -1340000e^{\frac{1}{2000}t}-6400t+h\end{array}$

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

$0=-1340000e^0-6400\left(0\right)+h$

Then we have

$h=1340000$

After that, substitute with the value of constantinto equation (b), then we have

$s\left(t\right)=-6400t-1340000e^{-\frac{1}{2200}t}+1340000$

is the distance for the cannonball from the shotting point at time t.

Find the maximum height

Now, we have to obtain the maximum height that the cannonball reaches as the following:

First, we have to obtain the time at which the velocity of cannonball is equal to 0 by substituting with$v=0$ into equation (3) as

$\begin{array}{rcl}0& =& 6700e^{\frac{1}{200}t}-6400\\ 6700e^{-\frac{1}{200}t}& =& 6400\\ e^{\frac{1}{200}}t& =& \frac{64}{67}\\ \ln e^{\frac{1}{200}t}& =& \ln \left(\frac{64}{67}\right)\\ -\frac{1}{200}t& =& \ln \left(\frac{64}{67}\right)\end{array}$

Then we have

$\begin{array}{rcl}t& =& -200\ln \left(\frac{64}{67}\right)\\ & =& 9.16\text{seconds}\end{array}$

Second, substitute with$t=9.16$ seconds into equation (c), then we have

$\begin{array}{rcl}s& =& -6400(9.16)-1340000e^{\frac{1}{200}(19.16)}+1340000\\ & =& -58636.2-1340000\left(\frac{64}{67}\right)+1340000\\ & =& 1363.8\text{ft}\end{array}$

which is less than that in problem (36).

$$\begin{gathered} v\left(t\right)=6700e^{-\frac{1}{2000}t}-6400 \\ s\left(t\right)=-6400t-1340000e^{\frac{1}{200}t}+1340000 \end{gathered}$$

Maximum height$=1363.8\text{ft}$