A small block of mass 100 g is pushed against a horizontal spring...

By the conservation of mechanical energy, the elastic potential energy of the spring is converted into the kinetic energy of the block when it detaches from the spring at its natural length:

$\dfrac{1}{2} k x^2 = \dfrac{1}{2} m v^2$

$v = x \sqrt{\dfrac{k}{m}}$

Substituting the given values $k = 100 \text{ N/m}$, $m = 100 \text{ g} = 0.1 \text{ kg}$, and $x = 5.0 \text{ cm} = 0.05 \text{ m}$:

$v = 0.05 \sqrt{\dfrac{100}{0.1}} = 0.05 \sqrt{1000} = 0.5\sqrt{10} \text{ m/s}$

The block travels horizontally and falls a vertical distance $h = 2 \text{ m}$. The time of flight $t$ is given by:

$h = \dfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}}$

Taking $g = 10 \text{ m/s}^2$:

$t = \sqrt{\dfrac{2 \times 2}{10}} = \sqrt{0.4} = \dfrac{2}{\sqrt{10}} \text{ s}$

The horizontal distance traveled by the block from the free end of the spring is:

$R = v \times t = 0.5\sqrt{10} \times \dfrac{2}{\sqrt{10}} = 1 \text{ m}$