Question

Figure (8-E4) shows a particle sliding on a frictionless.; track which terminates in a straight horizontal section.. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground ?

Answer 1

$H=1m,h=0.5m$

Applying law of conservation of Energy for point A and B

$mgH=\frac{1}{2}m{v}^2+mgh$

$\Rightarrow g-\left(\frac{1}{2}\right)v^2=0.5g$

$\Rightarrow v^2=2(g-0.5g)=g$

$\Rightarrow v=\sqrt{g}=3.1m/s$

After point B the body exhibits projectile motion for which,

$\theta =0$

$y=-0.5$

So, $-0.5=\left(usin\theta \right)t-\left(\frac{1}{2}\right)g{t}^2$

$\Rightarrow 0.5=44.9\times t^2$

$\Rightarrow t=0.31sec$

So, $x=\left(vcos\theta \right)t$

$3.1\times 0.31=1m$

So, the particle will hit the ground at a horizontal distance m from B.