A body is thrown vertically upwards. If air r | vedclass

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(B) Let the initial velocity of the body be $u$ and the acceleration due to air resistance be $a$.During the upward motion,both gravity $(g)$ and air

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Less than the time of fall

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(B) Let the initial velocity of the body be $u$ and the acceleration due to air resistance be $a$.During the upward motion,both gravity $(g)$ and air resistance $(a)$ act downwards. The net acceleration is $(g + a)$. The time of rise $t_1$ is given by $t_1 = \frac{u}{g + a}$.The maximum height reached is $H = \frac{u^2}{2(g + a)}$.During the downward motion,gravity acts downwards while air resistance acts upwards. The net acceleration is $(g - a)$. The time of fall $t_2$ is given by $H = \frac{1}{2}(g - a)t_2^2$.Substituting $H$,we get $\frac{1}{2}(g - a)t_2^2 = \frac{u^2}{2(g + a)}$,which simplifies to $t_2 = \frac{u}{\sqrt{(g + a)(g - a)}}$.Comparing $t_1$ and $t_2$,we see that $t_2 = t_1 \sqrt{\frac{g + a}{g - a}}$. Since $\sqrt{\frac{g + a}{g - a}} > 1$,it follows that $t_2 > t_1$. Thus,the time of rise is less than the time of fall.

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

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