Q3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

If the object in Problem 1 has a mass of  500 kg instead of 5 kg , when will it strike the ground? [Hint: Here the exponential term is too large to ignore. Use Newton’s method to approximate the time t when the object strikes the ground (see Appendix B)

Step-by-Step Solution

Answer

The equation of motion of the object is  $x\left(t\right)=0.981t+0.981e^{\frac{-t}{10}}-981m$.The takes by the object to strike the ground is $\mathbf{18}\mathbf{.}\mathbf{6631} \mathbf{Sec}\mathbf{.}$

Step 1: Important concept.

Use Newton’s method to approximate the time t when the object strikes the ground 

$t_{n+1}=t_n-\frac{f\left(t_n\right)}{f\text{'}\left(t_n\right)}$

Step 2: Find the equation of motion of an object

The given values are $m=500$,  $v_0=0$,  $g=9.81$,   $v_0=0$,  $b=50$,

The equation of motion is  $x\left(t\right)=\frac{mgt}{b}+\frac{m}{b}\left(v-\frac{mg}{b}\right)\left(1-e^{\frac{-bt}{m}}\right) ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Put all the given values in (1)

 $\frac{500\left(9.81\right)t}{50}+\frac{500}{50}\left(0-\frac{500\left(9.81\right)}{50}\right)\left(1-e^{\frac{-50t}{500}}\right)$

  $\mathbf{x} \left(\mathbf{t}\right)\mathbf{=}\mathbf{98}\mathbf{.}\mathbf{1} \mathbf{t}\mathbf{+}\mathbf{981} {\mathbf{e}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{10}}}\mathbf{-}\mathbf{981} \mathbf{m}$

 

Step 3: Find the result for what happens when object strike the ground when x   ( t ) = 1000   m

Put the value of $x\left(t\right)=1000 m$ then

 $$\begin{gathered} 1000=98.1t+981e^{\frac{-t}{10}}-981m \\ 1000+981=98.1t+981e^{\frac{-t}{10}} \\ 1981=98.1t+981e^{\frac{{\displaystyle -t}}{{\displaystyle 10}}} \\ t=\frac{1981}{98.1} \\ t=20.2 \end{gathered}$$

 

(Ignoring the exponential term is too large)

 

Therefore, the time  $\mathbf{t}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{2} \mathbf{Sec}\mathbf{.}$

Step 3: Apply Newton’s method

Let  $f\left(t\right)=1981=98.1t+981e^{\frac{-t}{10}}=0$.

The Newton’s method is  $t_{n+1}=t_n-\frac{f\left(t_n\right)}{f\text{'}\left(t_n\right)}$.

 $f\left(t\right)=98.1-98.1e^{\frac{-t}{10}}=\left(1-e^{\frac{-t}{10}}\right)98.1$

And the formula is $t_{n+1}=t_n-\frac{f\left(t_n\right)}{f\text{'}\left(t_n\right)}$ .

Put the value is  $n=0$,  $t_1=\frac{20.1936-\left(20.1936+10e^{\left(-2.011936\right)}-20.1936\right)}{\left(1-e^{-2.011936}\right)}$.

 $\left(t_0=\frac{1981}{98.1}=20.1936\right)$

 

Therefore, the result is ${\mathbf{t}}_{\mathbf{1}}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{663121} \mathbf{Sec}\mathbf{.}$