A

To determine the ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd, and 4th seconds, we can use the formula for the distance fallen under gravity, which is given by:
d = ut + \frac{1}{2}gt^2
where u is the initial velocity (0 for free fall), g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time in seconds.
1st second:
Distance fallen = \frac{1}{2}g(1)^2 = \frac{1}{2}g \approx 4.9 \, m
2nd second:
Distance fallen = \frac{1}{2}g(2)^2 - \frac{1}{2}g(1)^2 = 4g \approx 19.6 - 4.9 = 14.7 \, m
3rd second:
Distance fallen = \frac{1}{2}g(3)^2 - \frac{1}{2}g(2)^2 = 9.8g - 4g \approx 44.1 - 19.6 = 24.5 \, m
4th second:
Distance fallen = \frac{1}{2}g(4)^2 - \frac{1}{2}g(3)^2 = 16g - 9g \approx 78.4 - 44.1 = 34.3 \, m
Therefore, the distances travelled in the 1st, 2nd, 3rd, and 4th seconds are in the ratio of 1:4:9:16.