This problem involves the principles of conservation of energy for a particle falling under gravity. We need to find the height at which the particle's speed is half of its final speed (just before hitting the ground) and also determine the ratio of its kinetic energy to potential energy at that specific height.

Calculating Speed at Ground Level

Let's denote the initial height from which the particle is released as $S$. The particle starts from rest, so its initial kinetic energy (KE$_i$) is 0. Its initial potential energy (PE$_i$) relative to the ground (height = 0) is $m \times g \times S$, where $m$ is the mass of the particle and $g$ is the acceleration due to gravity.

Using the conservation of mechanical energy, the total energy at the release point is equal to the total energy just before hitting the ground.

Total Initial Energy = PE$_i$ + KE$_i$ = $mgS + 0 = mgS$.

Just before hitting the ground (height = 0), the potential energy (PE$_f$) is 0. Let the speed be $v_g$. The kinetic energy (KE$_f$) is $\frac{1}{2}mv_g^2$.

Total Final Energy = PE$_f$ + KE$_f$ = $0 + \frac{1}{2}mv_g^2$.

Equating initial and final energies:

$ mgS = \frac{1}{2}mv_g^2 $

From this, we can find the square of the speed just before hitting the ground:

$ v_g^2 = 2gS $

So, the speed is $v_g = \sqrt{2gS}$.

Finding the Specific Height

The problem states that at a certain height, let's call it $h$, the particle's speed ($v$) is half the speed it would have just before hitting the ground ($v_g$).

Therefore, $v = \frac{1}{2} v_g$.

Squaring both sides, we get:

$ v^2 = \left(\frac{1}{2} v_g\right)^2 = \frac{1}{4} v_g^2 $

Substituting the expression for $v_g^2$:

$ v^2 = \frac{1}{4} (2gS) = \frac{gS}{2} $

Now, let's apply the conservation of energy between the release point (height $S$) and this specific height $h$.

Energy at height $S$ = Energy at height $h$.

Energy at height $h$ = Potential Energy at $h$ + Kinetic Energy at $h$.

PE$_h = mgh$.

KE$_h = \frac{1}{2}mv^2 = \frac{1}{2}m \left( \frac{gS}{2} \right) = \frac{mgS}{4}$.

So, the total energy at height $h$ is:

$ E_h = mgh + \frac{mgS}{4} $

Equating the total energy at height $S$ ($mgS$) with the total energy at height $h$ ($E_h$):

$ mgS = mgh + \frac{mgS}{4} $

We can cancel out $m$ and $g$ from the equation (assuming $m \neq 0$ and $g \neq 0$):

$ S = h + \frac{S}{4} $

Now, solve for $h$:

$ h = S - \frac{S}{4} $ $ h = \frac{4S - S}{4} $ $ h = \frac{3S}{4} $

So, the height from the surface of the Earth is $\frac{3S}{4}$.

Determining the Energy Ratio

We need to find the ratio of the kinetic energy to the potential energy at this height $h = \frac{3S}{4}$.

We already calculated the kinetic energy at height $h$:

$ \text{KE}_h = \frac{mgS}{4} $

Now, let's calculate the potential energy at height $h$:

$ \text{PE}_h = mgh = mg \left( \frac{3S}{4} \right) = \frac{3mgS}{4} $

The ratio of kinetic energy to potential energy is:

$ \frac{\text{KE}_h}{\text{PE}_h} = \frac{\frac{mgS}{4}}{\frac{3mgS}{4}} $

Canceling the common terms ($\frac{mgS}{4}$):

$ \frac{\text{KE}_h}{\text{PE}_h} = \frac{1}{3} $

Conclusion

The height from the surface of the Earth is $\frac{3S}{4}$, and the ratio of the particle's kinetic energy to its potential energy at that instant is $\frac{1}{3}$.