A stone is falling freely and describes a distance `s` in `t` seconds given by equation `s=1/2"gt"^(2)`.
The acceleration of the stone is

Options

A. uniform
B. zero
C. non-uniform
D. indeterminate

Explanation

To find the acceleration of a stone falling freely, we start with the given equation for the distance \( s \) described by the stone in \( t \) seconds:

\[ s = \frac{1}{2} g t^2 \]

where \( g \) is the acceleration due to gravity.

### Step 1: Differentiate \( s \) with respect to \( t \) to find the velocity \( v \).

The first derivative of \( s \) with respect to \( t \) gives us the velocity \( v \):

\[ v = \frac{ds}{dt} = \frac{d}{dt} \left( \frac{1}{2} g t^2 \right) \]

Using the power rule of differentiation:

\[ v = \frac{1}{2} g \cdot 2t = gt \]

### Step 2: Differentiate \( v \) with respect to \( t \) to find the acceleration \( a \).

Now, we differentiate \( v \) with respect to \( t \):

\[ a = \frac{dv}{dt} = \frac{d}{dt} (gt) \]

Since \( g \) is a constant, we have:

\[ a = g \]

### Conclusion

The acceleration of the stone is:

\[ \boxed{g} \]