Distance Travelled by a Falling Body
Distance travelled by a body falling freely starting from rest can be calculated using the equation:
\[ s = \frac{1}{2}gt^2 \]
where:
- \( s \) is the distance travelled
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s\(^2\))
- \( t \) is the time in seconds

Ratio of Distances Travelled in 1st, 2nd, and 3rd Seconds
To find the ratio of distances travelled in 1st, 2nd, and 3rd seconds, we can substitute the respective time values into the equation.
- Distance travelled in 1st second:
\[ s_1 = \frac{1}{2} \times 9.81 \times 1^2 = 4.905 \text{ m} \]
- Distance travelled in 2nd second:
\[ s_2 = \frac{1}{2} \times 9.81 \times 2^2 = 19.62 \text{ m} \]
- Distance travelled in 3rd second:
\[ s_3 = \frac{1}{2} \times 9.81 \times 3^2 = 44.145 \text{ m} \]

Ratio Calculation
Now, we can find the ratio of distances travelled in 1st, 2nd, and 3rd seconds:
\[ \frac{s_1}{s_2} = \frac{4.905}{19.62} \approx 1:4 \]
\[ \frac{s_2}{s_3} = \frac{19.62}{44.145} \approx 1:2.25 \]
Therefore, the ratio of distances travelled by a falling body starting from rest in the 1st, 2nd, and 3rd seconds are approximately 1:4 and 1:2.25 respectively.

No... the answer will be 1:3:5 in the ratio...