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?

Short 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-by-step solution

Find the weight of the object

For finding the weight of the object apply:

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

role="math" localid="1664169361222" $$\begin{gathered} ma=W-10v \\ m\frac{dv}{dt}=400-10v \\ W=mg\left(a=\frac{dv}{dt}\right) \\ 400=32m \\ m=12.45kg \end{gathered}$$

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$C=-40$

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

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$c=450$

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

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$13.75 Sec.$