Question

A ball of mass M thrown upward if the air resistance is considered constant (R) What will be times of Ascent and decent. Give Mathematical proof.

Answer 1

Let M thrown upward with the velocity V

So deacceleration is +(g+R)

$v^2=u^2-2(g+R)S$

$S=\frac{u^2}{2(g+R)}$

$S=ut-\frac{1}{2}(g+R)t^2$

$\frac{1}{2}(g+R)t^2-ut+s=0$

$t=\frac{u\pm \sqrt{u^2-2(g+R)S}}{(g+R)}$

$t=\frac{u\pm \sqrt{v^2-u^2}}{g+R}$

$t=d\frac{u}{g-R}$

Which coming down acceleration is $=(g-R)$

$S=\frac{1}{2}(g-R)t^2$

$t^2=\frac{u^2}{(g^2-R^2)}$

$t=\sqrt{\frac{u^2}{(g^2-R^2)}}$