An object is released from rest and falls a distance $h$ during the first second of time. How far will it fall during the next second of time?

$ \textbf{(A)}\ h\qquad\textbf{(B)}\ 2h \qquad\textbf{(C)}\ 3h \qquad\textbf{(D)}\ 4h\qquad\textbf{(E)}\ h^2 $
1. **Determine the distance fallen in the first second:**
   The object is released from rest, so the initial velocity \( v_0 = 0 \). The distance fallen under constant acceleration \( a \) (gravity, \( g = 9.81 \, \text{m/s}^2 \)) in time \( t \) is given by:
   \[
   \Delta x = v_0 t + \frac{1}{2} a t^2
   \]
   Since \( v_0 = 0 \), the equation simplifies to:
   \[
   \Delta x = \frac{1}{2} a t^2
   \]
   For the first second (\( t = 1 \) s):
   \[
   h = \frac{1}{2} g (1)^2 = \frac{1}{2} g
   \]

2. **Determine the total distance fallen in the first two seconds:**
   Using the same formula for \( t = 2 \) s:
   \[
   \Delta x = \frac{1}{2} g (2)^2 = \frac{1}{2} g \cdot 4 = 2g
   \]

3. **Calculate the distance fallen during the second second:**
   The distance fallen during the second second is the total distance fallen in two seconds minus the distance fallen in the first second:
   \[
   \Delta x_{\text{second second}} = \Delta x_{\text{total 2 seconds}} - \Delta x_{\text{first second}}
   \]
   Substituting the values:
   \[
   \Delta x_{\text{second second}} = 2g - \frac{1}{2} g = \frac{4g}{2} - \frac{1g}{2} = \frac{3g}{2}
   \]

4. **Relate the distances to \( h \):**
   From step 1, we know \( h = \frac{1}{2} g \). Therefore:
   \[
   \Delta x_{\text{second second}} = \frac{3g}{2} = 3 \left( \frac{1}{2} g \right) = 3h
   \]

Conclusion:
\[
\boxed{\textbf{(C)}\ 3h}
\]