Question

A small ball of mass $m$ is thrown upward with velocity $u$ from the ground. The ball experiences a resistive force $mk{v}^2$ where $v$ is its speed. The maximum height attained by the ball is:

• A. $\frac{1}{k}{\tan }^{-1}\frac{k{u}^2}{2g}$
• B. $\frac{1}{2k}\ln \left(1+\frac{k{u}^2}{g}\right)$
• C. $\frac{1}{2k}\ln \left(1+\frac{k{u}^2}{2g}\right)$
• D. $\frac{1}{2k}{\tan }^{-1}\frac{k{u}^2}{g}$

Answer 1

The correct option is B

$\frac{1}{2k}\ln \left(1+\frac{k{u}^2}{g}\right)$

Step 1: Given data and drawing the diagram

Mass $=m$

Velocity $=u$

Resistive force $=mk{v}^2$

Acceleration $=a$

Net force $=F_{net}$

Speed $=v$

Step 2: Find maximum height attained by the ball

Net force on ball $=$ weight of ball $+$ resistive force

$\begin{array}{rcl}{F}_{net}& =& w+F_{resistive}\\ & =& (-mg)+(-mk{v}^2)\\ & =& -m(g+k{v}^2)\end{array}$

So, net acceleration of ball,

$\begin{array}{rcl}a& =& \frac{F_{net}}{m}\\ & =& \frac{-m\left(g+k{v}^2\right)}{m}\\ & =& -\left(g+k{v}^2\right)\end{array}$

Now,

$v\frac{dv}{ds}=-(g+k{v}^2)$ $\left[\because a=\frac{dv}{dt};v=\frac{ds}{dt}\right]$

$\Rightarrow$$\frac{vdv}{(g+k{v}^2)}=-ds$

Integration both sides,

${\int }_u^0\frac{vdv}{(g+k{v}^2)}dv=-{\int }_0^{H}ds$ ….(1)

Let $g+k{v}^2=t$

$\Rightarrow$$0+2kvdv=dt$

$\Rightarrow$ $vdv=\frac{1}{2k}dt$

Lower limit:

If $v=u$, then $g+k{u}^2=t$

$t=g+k{u}^2$

Upper limit:

If $v=0$, then $g+k{\left(0\right)}^2=t$

$t=g$

Putting these values in equation (1), we get

$\Rightarrow$ ${\int }_{g+k{u}^2}^g\frac{{\displaystyle \frac{1}{2k}}dt}{t}=-{\int }_0^{H}ds$

$\Rightarrow$ $\frac{1}{2k}{\int }_{g+k{u}^2}^g\frac{{\displaystyle dt}}{t}=-{\left[s\right]}_0^H$

$\Rightarrow$ $\frac{1}{2k}{\left[\ln \left(t\right)\right]}_{g+k{u}^2}^g=-\left[H-0\right]$

$\Rightarrow$ $\frac{1}{2k}\left[\ln \left(g\right)-\ln \left(g+k{u}^2\right)\right]=-\left[H\right]$

$\Rightarrow$$-\frac{1}{2k}\left[\ln \left(g\right)-\ln \left(g+k{u}^2\right)\right]=H$

$\Rightarrow$ $\begin{array}{rcl}H& =& \frac{1}{2k}\left[\ln \left(g+k{u}^2\right)-\ln \left(g\right)\right]\\ & =& \frac{1}{2k}\left[\ln \left(\frac{g+k{u}^2}{g}\right)\right]\\ & =& \frac{1}{2k}\ln \left(1+\frac{k{u}^2}{g}\right)\end{array}$

Hence, option B is correct.