AIIMS 2018 Physics Equations of Motion MCQ Question

To find the distance covered by the second drop in 0.5 seconds, we first note that the second drop is released 0.5 seconds after the first. By the time the second drop falls for 0.5 seconds, the first drop has already fallen for 1 second, covering a distance of $S_1 = \frac{1}{2} g t^2 = \frac{1}{2} (10 \, \text{m/s}^2)(1^2) = 5 \, \text{m}$. In the next 0.5 seconds, the second drop falls $S_2 = \frac{1}{2} g (0.5)^2 = \frac{1}{2} (10)(0.25) = 1.25 \, \text{m}$, but the first drop continues to fall, covering an additional $S_1' = \frac{1}{2} g (1.5)^2 = \frac{1}{2} (10)(2.25) = 11.25 \, \text{m}$. Thus, the total distance from the starting point for the second drop after 1 second is $5 + 1.25 = 6.25 \, \text{m}$, while the distance covered by the first drop in the same time is $ 11.25 , \text{