Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An object of mass 8 kg  is given an upward initial velocity of 20 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is -16v , where v is the velocity of the object in m/sec. Determine the equation of motion of the object. If the object is initially 100 m above the ground, determine when the object will strike the ground.

Step-by-Step Solution

Answer

The equation of motion of an object  $x\left(t\right)=4.9t-12.45e^{-2t}+112.45$ and the time takes by the object to strike the ground is 22.9 sec.

Step 1: Important hint.

Use Newton’s method to solve for t.

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

Step 2: Find the velocity

For finding the weight of the object apply:

 

Net force=W-drag force

 $$\begin{gathered} ma=W-16v \\ ma=mg-16v \\ 8\frac{dv}{dt}=8(9.81)-16v \\ 8v\text{'}=78.4-16v \end{gathered}$$

 Further solve the above expression,

 

  $$\begin{gathered} v\text{'}=9.8-2v \left(\text{Integrating factor} e^{2t}\right) \\ v\text{'}+2v=9.8 \\ v.e^{2t}=\int 9.8e^{2t}dt \\ v.e^{2t}=4.9e^{2t}+C \\ v=4.9e^{2t}+C{e}^{-2t} \\ v.e^{2t}=4.9e^{2t}+C \end{gathered}$$

 

 At the value of  $v=-20 and t=0, then C=24.9$ (c is constant and arrange negative sign)

 $v=4.9+24.9e^{-2t}$

 

Step 3: Find the equation of motion

$$\begin{gathered} v=\frac{dx}{dt} \\ \frac{dx}{dt}=4.9+24.9e^{-2t} \\ x\left(t\right)=4.9t-12.45e^{-2t}+c \end{gathered}$$

Put the value of  $t=0, x=100, then C=112.45$

 $x\left(t\right)=4.9 t-12.45 e^{-2t}+112.45$

 

Step 4: Find the value of t

When the value of  $x=0$

 $0=4.9t-12.45e^{-2t}+112.45$

 By trial and error, the value of t  $t=22.9\sec .$

 Therefore, the value of t is $t=22.9\sec .$