As depicted in the figure, a particle slides on a frictionless...

Let the initial height of the particle be $h_1 = 1.0\text{ m}$ and the height of the horizontal section of the track be $h_2 = 0.5\text{ m}$.

By the principle of conservation of mechanical energy, the velocity $v$ of the particle at the end of the track is given by:

$mg h_1 = mg h_2 + \dfrac{1}{2} m v^2$

$v = \sqrt{2g(h_1 - h_2)}$

Substituting the given values:

$v = \sqrt{2g(1.0 - 0.5)} = \sqrt{g}$

The particle leaves the track horizontally. The time $t$ taken to fall to the ground from height $h_2$ is:

$t = \sqrt{\dfrac{2 h_2}{g}} = \sqrt{\dfrac{2(0.5)}{g}} = \dfrac{1}{\sqrt{g}}$

The horizontal distance $x$ traveled by the particle from the edge of the track is:

$x = v \times t = \sqrt{g} \times \dfrac{1}{\sqrt{g}} = 1.0\text{ m}$