Question

let a be the time it takes the cannonball to attain its maximum height and let td be the time it takes the cannonball to fall from the maximum height to the ground. Compare the value of t with the value of td and compare the magnitude of the impact velocity v with the initial velocity v_0. See Problem 50 in Exercises 3.1. A root finding application of a CAS might be useful here. [Hint: Use the model in Problem 15 when the cannonball is falling.]

Short answer

The value of t with the value of td and compare the magnitude of the impact velocity v with the initial velocity Which is lower than the initial velocity.$\left(v_0=300\text{ft}/\text{s}\right)>\left(v_i=207.4\text{ft}/\text{s}\right)$

Step-by-step solution

 Step 1: Define the newton law of differential equation

Differential equations are used in research and engineering to model physical quantities that fluctuate over time.

Newton's law, which is a second-order differential equation$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}\mathbf{=}\mathit{m}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{{\mathbf{dt}}^{\mathbf{2}}}$, is the most famous example. This equation represents a moving object's position x(t) as a function of time.

Find the maximum height can be attained by the cannonball

Ny using newton law$m\frac{dv}{dt}=-mg-k{v}^2$

When$k=0.03$

$v(t=0)=300\text{ft}/\text{s}$

Use second order differential equation to solve

$\begin{array}{rcl}& & \frac{m}{k}\int \frac{dv}{-\frac{mg}{k}-v^2}\\ & =& \int dt\frac{0.5}{0.0003}\int \frac{dv}{\left(-53333.33-v^2\right)}\\ & =& \int dt\int \frac{dv}{53333.33+v^2}\\ & =& -\frac{3}{5000}t+c_1\end{array}$

Integrate by substitution method.

$$\begin{gathered} u=\frac{v}{\sqrt{53333.33}}\text{,} \\ du=\frac{1}{\sqrt{53333.33}}dt \end{gathered}$$

$\int \frac{1}{53333.33+v^2}dt=\int \frac{\sqrt{53333.33}}{53333.33+53333.33u^2}du$

$$\begin{gathered} =\frac{\sqrt{53333.33}}{53333.33}\int \frac{1}{1+u^2}du \\ =\frac{1}{\sqrt{53333.33}}{\tan }^{-1}\left(u\right) \\ =\frac{1}{\sqrt{53333.33}}{\tan }^{-1}\left(u\right) \\ =\frac{1}{\sqrt{53333.33}}{\tan }^{-1}\left(\frac{v}{\sqrt{53333.33}}\right) \\ \frac{1}{\sqrt{53333.33}}{\tan }^{-1}\left(\frac{v}{\sqrt{53333.33}}\right)=-\frac{3}{5000}t+c_1 \\ \frac{v}{\sqrt{53333.33}}=\tan \left(-0.1386t+c_2\right) \\ v=\sqrt{53333.33}\tan \left(-0.1386t+c_2\right) \end{gathered}$$

Now find the constant C2 by applying (v, t) = (300ft/sec)

$$\begin{gathered} 300=\sqrt{53333.33}\tan \left(0+c_2\right) \\ \tan \left(c_2\right)=1.3 \end{gathered}$$

Then we have

$$\begin{gathered} c_2={\tan }^{-1}(1.3) \\ =52.4 \end{gathered}$$

Now find C3

$v\left(t\right)=\sqrt{53333.33}\tan (52.4-1.386t)$is the velocity of cannonball.

Now find the height of the cannonball.

$$\begin{gathered} \int ds=\int vdt \\ s=\sqrt{53333.33}\int \tan (52.4-1.386t)dt \\ =\sqrt{53333.33}\frac{500\ln \left(\cos \left[\frac{693x-26200}{500}\right]\right)}{693}+c \\ =166.62\ln \left(\cos \left[\frac{693x-26200}{500}\right]\right) \end{gathered}$$

Now find the constant C by applying (s, t) = (0,0)

$0=166.62\ln \left(\cos \left[\frac{0-26200}{500}\right]\right)+c$

Then we have $0=166.62\ln (0.61)+c$

$$\begin{gathered} c=-166.62\ln (0.61) \\ \text{= 82.3} \end{gathered}$$

Now substitute the value of c in the equation

$s\left(t\right)=166.62\ln \left(\cos \left[\frac{693t-26200}{500}\right]\right)+82.3$

When $v=0$

$$\begin{gathered} 0=\sqrt{53333.33}\tan \left(52.4-1.386t_a\right) \\ 52.4-1.386t_a={\tan }^{-1}\left(0\right) \\ {\text{52.4 = 1.386t}}_a \\ {t}_a=\frac{{\displaystyle 52.4}}{{\displaystyle 1.386}} \\ =37.8\text{seconds} \end{gathered}$$

Find the time that the cannonball takes to reach the ground from its shotting by substituting with .

$$\begin{gathered} 0=166.62\ln \left(\cos \left[\frac{693t-26200}{500}\right]\right)+82.3 \\ -82.3=166.62\ln \left(\cos \left[\frac{693t-26200}{500}\right]\right)-0.494=\ln \left(\cos \left[\frac{693t-26200}{500}\right]\right) \\ \cos \left[\frac{693t-26200}{500}\right]=e^{-0.494} \\ \frac{693t-26200}{500}={\cos }^{-1}(0.61) \\ \text{693 t - 26200 = 26197.35} \\ \text{693 t = 26200 + 26197.35} \\ t\approx 75.6\text{seconds} \end{gathered}$$

we can obtain the time $t_d$at which the cannonball reaches the ground from its maximum height as ${\text{t}}_d{\text{= t -t}}_a$

$$\begin{gathered} \text{= 75.6 - 37.8} \\ =37.8\text{seconds} \end{gathered}$$

which is the same as the time at which the cannonball reaches its maximum height.

Find the initial velocity of the cannonball.

When$t=0$substitute in the equation

$\begin{array}{rcl}{v}_0& =& \sqrt{53333.33}\tan (52.4-0)\\ & =& 230.94\times 1.3\\ & =& 230.94\times 0.898\\ & =& 300\text{ft}/\text{s}\end{array}$

Now find the impact velocity of the cannonball when $t=75.6$

$v_i=\sqrt{53333.33}\tan (52.4-0.1386\times 75.6)$