Question

An object falling from rest in air is subjected not only to the gravitational force but also air resistance which is proportional to the velocity with constant of proportionality as $k>0,$ and acts in a direction opposite to motion $(g=9.8m/s^2).$ Then the velocity cannot exceed:

• A. $\frac{9.8}{k}m/s$
• B. $\frac{98}{k}m/s$
• C. $\frac{k^2+3}{9.8}m/s$
• D. $\frac{k^2-3}{9.8}m/s$

Answer 1

The correct option is A $\frac{9.8}{k}m/s$

Let $V\left(t\right)$ be the velocity of the object at time $t$.
Given $\frac{dV}{dt}=9.8-kV\Rightarrow \frac{dV}{9.8-kV}=dt.$
Integrating, we get $\ln |9.8-kV|=-kt+\ln |c|$
$9.8-kV=C{e}^{-kt}$
But $V\left(0\right)=0\Rightarrow C=9.8$
Thus, $9.8-kV=9.8e^{-kt}$
$\Rightarrow kV=9.8(1-e^{-kt})$
$\Rightarrow V\left(t\right)=\frac{9.8}{k}(1-e^{-kt})<\frac{9.8}{k}\forall t.$ Hence, $V\left(t\right)$ cannot exceed $\frac{9.8}{k}m/s$.