Question

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10m$. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point $C$ from point $A$ is $t(\sqrt{2}+1)$ s. The value of $t$ is. _____(use $g=10m/s^2$ )

Answer 1

Answer: 2

From E.C. $=\frac{1}{2}m{v}_0^2=mgh$
$v_0=10\sqrt{2}$
For $A\to B$
at $B,v=0$
$a=-g\sin 45^{\circ }=\frac{-10}{\sqrt{2}}$
$v=u+a{t}_1\Rightarrow 0=10\sqrt{2}-\frac{10}{\sqrt{2}}{t}_1\Rightarrow t_1=2sec$
For $B\to C$
$s=u{t}_2+\frac{1}{2}a{t}_2^2$
$\frac{10}{\sin 30^{\circ }}=\frac{1}{2}\left(10\sin 30^{\circ }\right)t_2^2$
$t_2=2\sqrt{2}$
So total time
$T=t_1+t_2$
$=2\sqrt{2}+2$
$=2(\sqrt{2}+1)sec$