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

Answer 1

The correct option is C Less than the time of fall

Let the initial velocity of ball $=u$
and the air resistance has acceleration$=a$
So, while going up,
net acceleration in downward direction$=g+a$
and while coming down,
net acceleration in downward direction$=g-a$

For upward journey,
By using equations of motion, we get
Time of rise, $t_1=\frac{u}{g+a}$
and height reached $H=\frac{u^2}{2(g+a)}$

For downward motion,
Time of fall $t_2$ can be calculated using second equation of motion,
$H=\frac{u^2}{2(g+a)}=\frac{1}{2}(g-a)t_2^2$

$\Rightarrow t_2=\frac{u}{\sqrt{(g+a)(g-a)}}=\frac{u}{(g+a)}\sqrt{\frac{g+a}{g-a}}$

$\therefore t_2>t_1$ because $\frac{1}{g+a}<\frac{1}{g-a}$

Hence, option $\left(\text{B}\right)$ is the right choice.