A

Given:
Height, h, falls freely under gravity.
Time taken to reach the ground, t.
The distance covered in the first part of the fall can be expressed using the formula for distance under constant acceleration (gravity in this case).
Using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
- s is the distance fallen
- u is the initial velocity (0 for free fall from rest)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken

The distance covered in the first height h/16 is:
\[ h = \frac{1}{2} g t_1^2 \]
Setting s = h/16 and rearranging gives us:
\[ h/16 = \frac{1}{2} g t_1^2 \]

Therefore,
\[ t_1 = \sqrt{\frac{h}{8g}} \]

When the total time is given, t,
To find t1 in terms of t:
\[ t_1 = \frac{t}{4} \]
Hence the time taken to cover the first h/16 height is \( t/4 \).
Therefore, the answer is A.