Answer to Question #121696 in Classical Mechanics for Quanda

Question

A particle Q of mass m falls under gravity in a medium, whose resistance to motion is of magnitude mkv^2, where v is the speed of Q and k is a positive constant. Given that the maximum speed of Q is U, show that U=√(g/k) (g is the acceleration due to gravity).
Given that Q is projected vertically upwards with speed V(>U), show that the speed of Q is equal to U when the height of Q above the point of projection is U^2/2g ln⁡[1/2 (1+V^2/U^2 )].
Find in terms of U, V and g, the time taken for the speed of Q to decrease from V to U.

Answer 1

As per the given question,

The falling particle Q of mass is m, is experiencing a resistiv force $=mkv^2$

Here v is the speed and k is the positive constant

Now applying the force balance equation of the net resultant force,

$\Rightarrow F=mg- mkv^2$

$\Rightarrow m \frac{dx^2}{dt^2}=m(g-kv^2)$

$\Rightarrow \frac{dx^2}{dt^2}=(g-kv^2)$

Now we can write this as,

$\Rightarrow \frac{d}{dt}(\frac{dx}{dt})=(g-kv^2)$

$\Rightarrow$ $\frac{dv}{dt}=g-kv^2$

Now applying the maxima and minima concept,

$\frac{dv}{dx}=0$

$\Rightarrow g-kv^2=0$

$\Rightarrow v=\sqrt{\frac{g}{k}}$

ii)

$\Rightarrow \frac{dv}{dt}=g-kv^2$

$\Rightarrow \frac{dv}{g-kv^2}=dt$

now after the integration,

$t=\frac{U^2}{2g} \ln⁡[\frac{1}{2 (1+V^2/U^2 )}].$