Q 3.4-2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A 400-lb object is released from rest 500 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is -10V, where v is the velocity of the object in ft/sec determine the equation of motion of the object. When will the object hit the ground?

Step-by-Step Solution

Answer

The equation of motion of the object is $x\left(t\right)=40t+50e^{-\left(0.8\right)t}-450$. The time takes the object hits the ground $13.75 Sec.$

Step 1: Find the weight of the object

For finding the weight of the object apply:

 $\text{Net force}=W-\text{drag force}$

$$\begin{gathered} ma=W-10v \\ m\frac{dv}{dt}=400-10v \\ W=mg \left(a=\frac{dv}{dt}\right) \\ 400=32m \\ m=12.45kg \end{gathered}$$

     

Step 2: Find the velocity

Apply formula for finding the velocity:

$$\begin{gathered} ma=W-10v \\ m\frac{dv}{dt}=400-10v \\ 12.5\frac{dv}{dt}=400-10v \\ \frac{dv}{dt}=32-0.8v \end{gathered}$$

Further solve the above expression:

$$\begin{gathered} v.e^{0.8t}=\int 32e^{0.8t}dt \\ v.e^{0.8t}=40e^{0.8t}+C \left(\text{Integrating factor} e^{0.8t}\right) \left(c=\text{integration constant}\right) \\ v=40+C{e}^{-0.8t} \end{gathered}$$ 

At $v=0, t=0$ then $C=-40$

$v=40-40e^{-0.8t}$

Step 3: Find the equation of motion

Now use value of for finding the value of $x\left(t\right)$.

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

$$\begin{gathered} \frac{dx}{dt}=40-40e^{-0.8t} \\ x\left(t\right)=40t+50e^{-0.8t}+C \end{gathered}$$

Put the value of $t=0, x=500$, then $c=450$

 $x\left(t\right)=40t+50e^{-0.8t}-450$

Step 4: Find the value of t .

The value is $x\left(t\right)=40t+50e^{-0.8t}-450$.

When object hits the ground $x=0$.

$40t+50e^{-0.8t}=450$ 

By solving trial and error method $t=13.75 Sec.$

Therefore, the equation of motion of the object is $x\left(t\right)=40t+50e^{-\left(0.8\right)t}-450$. The time takes the object hits the ground $13.75 Sec.$