A

Step 1: When an object is thrown upwards with initial speed \( u \), the time \( t_1 \) to reach the highest point is given by the formula: \( t_1 = \frac{u}{g} \), where \( g \) is the acceleration due to gravity.
Step 2: The total time \( t \) taken to hit the ground can be expressed as \( t = t_1 + t_2 \), where \( t_2 \) is the time taken to fall back down. From kinematics, it can be shown that \( t_2 = t_1 \). Thus, we find that the total time taken is: \( t = 2t_1 = 2 \cdot \frac{u}{g} \).
Step 3: The relation with respect to heights indicates that \( H = \frac{u^2}{2g} \). This leads to the relationship between the times: \( t = 2t_1 = 2 \cdot \frac{u}{g} = n \cdot t_1 \). Therefore, eliminating \( n \), we derive: \( 2gH = n^2u^2 \).
Therefore, Option A is the correct choice.