Question

A block slides down a slope of angle $\theta$ with constant velocity. It is then projected up with a velocity of $10{ms}^{-1}$, $g=10{ms}^{-2}$ and $q={30}^o$. The maximum distance it can go up the plane before coming to stop is

• A. $10m$
• B. $5m$
• C. $4m$
• D. $15m$

Answer 1

Answer: (A)

$10m$

From above figure

for constant velocity

$mgsin{30}^{\circ }=\mu mgcos{30}^{\circ }$

$\mu =\frac{1}{\sqrt{3}}$

when it is thrown upward

$a=gsin{30}^{\circ }+\mu gcos{30}^{\circ }$

$=10M{\S }^2$

so in upward journey

u = 10m/s $a=-m/s^2$ v=0 s = ?

$v^2=u^2+2as$

$0=100-2\left(10\right)s$

$s=10m$

Hence (A) option is correct