91Ó°ÊÓ

Newton's Second Law of Motion states that the rate of change of instigation of a body is directly commensurable to the applied force and takes place in the direction in which the force acts.

It is Given that an object of mass mfalls from rest under gravity subject to an air resistance proportional to its speed.

Now the force that acts on that object is the gravitational force (acting downwards), that is$F_g=mg$ and the air resistance acting in the opposite direction of the motion (that is upward). Thus, it has been noticed that this force is proportional to the speed, so it can be written that, by using Newton's second law of motion we can write

$$\begin{gathered} ma=∑F \\ m\frac{dv}{dt}=mg-kv \end{gathered}$$

Now solve the differential equation by using the separation of variables method

$$\begin{gathered} ∫\frac{dv}{mg-kv}=\frac{1}{m}∫1dt \\ \frac{\ln \left(mg-kv\right)}{k}=\frac{t}{m}+C \\ mg-kv=C_2{e}^{-\frac{kt}{m}} \end{gathered}$$

So the general solution is

$v\left(t\right)=\frac{mg-C_2{e}^{-kt/m}}{k}$

To find the particular solution use the initial condition, that is v=0 when t=0 therefore$C_2=mg$

And the particular solution is

$v\left(t\right)=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}}\right)$

The terminal speed when $t→∞$is

$v\left(t→∞\right)=\frac{mg}{k}$

Now the terminal speed is the maximum speed the object is falling, therefore directly from the differential equation by making it equal to zero is

$$\begin{gathered} \frac{dv}{dt}=0=mg-kv \\ {v}_{\max }=\frac{mg}{k} \end{gathered}$$

In general, the value of the constant t is equal to the weight of the human divided by the terminal speed. On average, the terminal speed of the human body is $v_{\max }=53m/s$, and the average mass is$m=70kb$ Also the gravitational force near the earth's surface is $g=9.81m/s^2$,

Therefore, the value of the constant k is

When the drop reaches the terminal speed the air resistance force and the gravitational force are equal, therefore, the terminal speed is

$$\begin{gathered} k{v}_{\max }=mg \\ {v}_{\max }=\frac{mg}{k} \end{gathered}$$

And 99% of the terminal speed is

$99\%v_{\max }=0.99\frac{mg}{k}$

Assuming the drops starts their motion from rest, therefore use the particular solution

$$\begin{gathered} 0.99\frac{mg}{k}=\frac{mg}{k}\left(1-e^{-kt/m}\right) \\ 0.01=e^{-kt/m} \\ -\ln \left(100\right)=-\frac{k}{m}t \\ t=\frac{m}{k}\ln \left(100\right) \end{gathered}$$

Hence,the particular solution is$v\left(t\right)=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}}\right)$and the terminal speed is $v_{\max }=\frac{mg}{k}$

(a) When a person drops from an airplane with a parachute than a reasonable value of k is $k=12.96Kg/s$

(b) the time required for a drop starting at rest to reach 99% of its terminal speed is