Question: 7: The ratio of the distances travelled by a freely falling body in the 1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }} and 4^{\text {th }} second
(1) 1: 3: 5: 7
(2) 1: 1: 1: 1
(3) 1: 2: 3: 4
(4) 1: 4: 9: 16
Answer: Option (1)
Explanation:
For a freely falling body starting from rest, the acceleration is constant and equal to g.
The distance travelled in time t is given by
s = \frac{1}{2}gt^{2}.
The distance travelled in the n^{\text{th}} second is obtained by subtracting the
distance covered in (n-1) seconds from that covered in n seconds:
s_n = \frac{1}{2}g\left[n^{2} - (n-1)^{2}\right].
Simplifying,
s_n = \frac{1}{2}g(2n - 1).
This shows that the distances travelled in successive seconds are proportional to odd natural numbers.
Therefore, the ratio of distances travelled in the
1^{\text{st}}, 2^{\text{nd}}, 3^{\text{rd}} and 4^{\text{th}}
seconds is 1 : 3 : 5 : 7.