Question

A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is

• A. Equal to the time of fall
• B. Less than the time of fall
• C. Greater than the time of fall
• D. Twice the time of fall

Answer 1

Answer: (B)

Let the initial velocity of ball be $u$.
Time of rise $t_1 = \frac{u}{g + a} $ and height reached
$ h = \frac{u^2}{2( g + a)}$
Time of fall $t_2$ is given by
$\frac{1}{2} ( g - a) t_2^2 = \frac{u^2}{2(g + a)}$
$\Rightarrow t_2 = \frac{u}{\sqrt{(g + a)( g - a)}} $
$ = \frac{u}{(g + a)} \sqrt{\frac{ g + a}{g - a}}$
$\therefore t_2 > t_1$ because $\frac{1}{ g +a } < \frac{1}{g - a}$