Question

A ball of mass '$m$' is projected vertically upwards from point A on ground with initial velocity $u=60m/s$ as shown in the figure above. It is given that the fluid (above ground) exerts resistive force on the ball directly proportional to its velocity at that instant, in a direction opposite to the motion. Net force acting on the ball is $2mg$ downwards at the instant of projection. If time of ascent of ball in the fluid is $x\ln 2s$ ., then '$x$' is ($g=10m/s^2)$

• A. $6$
• B. $8$
• C. $5$
• D. $7$

Answer 1

The correct option is A $6$

Initially, force equation for particle is

$ku+mg=2mg$ where, $k=$ constant

So, $k=\frac{mg}{u}$

Let at any instant, speed of ball is $v$

So, $-a=g+\frac{kv}{m}$

$-a=g+\frac{gv}{u}$

$-\frac{dv}{dt}=g(1+\frac{v}{u})$

${\int }_u^0\frac{-dv}{g(1+\frac{v}{u})}={\int }_0^{t}dt$

$t=-\frac{u}{g}{\left[\ln (1+\frac{v}{u})\right]}_u^0$

$t=\frac{u\ln 2}{g}=\frac{60\ln 2}{10}$