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

The correct option is B

Less than the time of fall

Let the initial velocity of ball be u

Time of rise $t_1=\frac{u}{g+a}$ and height reached = $\frac{u^2}{2(g+a)}$

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 $\frac{1}{g+a}<\frac{1}{g-a}$