Question

An object falling from rest in air is subject not only to the gravitational force but also to air resistance. Assume that the air resistance is proportional to the velocity with constant of proportionality as $k>0$, and acts in a direction opposite to motion $(g=9.8m/S^2)$. If the velocity cannot exceed $A$ m/s then the value of $10Ak$ is

Answer 1

Let V(t) be the velocity of the object at time t.
We have $\frac{dV}{dt}=9.8-kV$

$\Rightarrow \frac{dV}{9.8-kV}=dt$.

Integrating, we have
$\log (9.8-kV)=-kt+C\Rightarrow 9.8-kV=const{e}^{-kt}$

But V(0) = 0 so const = 9.8. Thus, we have $9.8-kV=9.8e^{-kt}$

$\Rightarrow kV=9.8(1-e^{-kt})\Rightarrow V\left(t\right)=\frac{9.8}{k}(1-e^{-kt})<\frac{9.8}{k}$

for all t. Hence V(t) cannot exceed $\frac{9.8}{k}$. Thus 10 Ak is $98$.