Question

A particle is thrown upwards from ground. It experiences a constant air resistance force which can produce a retardation of $2 \,m/s^2$. The ratio of time of ascent to the time of descent is:

• A. $1 : 1$
• B. $\sqrt{\frac{2}{3}}$
• C. $\frac{2}{3}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

Answer: (B)

Let $a$ be the retardation produced by resistive force.
Also let $t_a$ and $t_d$ be the time ascent and descent respectively.
If the particle rises upto a height $h$
Then $h = \frac{1}{2} ( g + a) t^2_a$ and $ h = \frac{1}{2} (g - a) t^2_d$
$\therefore \frac{t_a}{t_d} = \sqrt{\frac{g - a}{g + a}} $
$= \sqrt{\frac{10 - 2}{10 + 2}} = \sqrt{\frac{2}{3}}$