Question

A body of mass $m$ projected vertically upwards with velocity and air resistance $k$ times the velocity is governed by the differential equation $\frac{dv}{dt}=-g-\frac{kv}{m}$.The maximum height attained by the body is?

• A. $\frac{m}{k}log(1+\frac{k}{mg}v)$
• B. $\frac{m}{k}log(1+\frac{k}{mg}{v}^2)$
• C. $\frac{m}{k^2}log(1+\frac{k}{mg}v)$
• D. $\frac{1}{k}log(1+\frac{k}{mg}v)$

Answer 1

Answer: (A)

$\frac{m}{k}log(1+\frac{k}{mg}v)$

We have $\frac{dv}{dt}=-g-\frac{kv}{m}$

This is a differential equation of variable separable type.

$\therefore \frac{dv}{g+\left(k/m\right)v}=-dt$

By integration, $\frac{m}{k}\log (g+\frac{k}{m}v)=-t+c$

When $t=0,v=V,\frac{m}{k}\log (g+\frac{k}{m}V)=c$

$\therefore t=\frac{m}{k}\log (g+\frac{k}{m}V)-\frac{m}{k}\log (g+\frac{k}{m}v)$

$=\frac{m}{k}\log \left(\frac{g+\left(k/m\right)V}{g+\left(k/m\right)v}\right)$

When the body attains maximum height, $v=0$.

$\therefore t=\frac{m}{k}\log \left(\frac{g+\left(k/m\right)V}{g}\right)=\frac{m}{k}\log (1+\frac{k}{mg}v)$.