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2.Motion in Straight Line
hard
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
Solution
(b) 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 $\frac{1}{{g + a}} < \frac{1}{{g – a}}$
Standard 11
Physics
