Question

A body of mass $m$ is lunched up an a rough indines plane making an angle ${45}^o$ with horizontal. If time of ascent is $\frac{1}{3}$ of time of descent, the frictional coefficient between plane and body is

Answer 1

$t_A=\sqrt{\frac{2S}{a_A}}$ & $t_D=\sqrt{\frac{2S}{a_D}}$
$t_A=\frac{1}{3}{t}_D$
$\sqrt{\frac{2S}{g\sin \theta +\mu \cos \theta }}=\frac{1}{3}\sqrt{\frac{2S}{g\sin \theta -\mu g\cos \theta }}$
$9g\sin \theta -9\mu g\cos \theta =g\sin \theta +\mu g\cos \theta$
$8g\sin \theta =10\mu g\cos \theta$
$\frac{8}{10}\frac{\sin \theta }{\cos \theta }=\mu$
$\therefore \mu =\frac{4}{5}\tan \theta$ $(\because \tan ={45}^o)$
So, $\mu =\frac{4}{5}$ = $0.8$