A body falls from a height $h=200 ext{ m}$ (at New Delhi). The ratio of distance travelled in each $2 ext{

Question

A body falls from a height $h=200 	ext{ m}$ (at New Delhi). The ratio of distance travelled in each $2 	ext{ sec}$ during $t = 0$ to $t = 6$ second of the journey is:

Accepted Answer

1. **Analyze the Motion:** The body falls from rest ($u=0$) under constant acceleration due to gravity ($a=g$). The problem asks for the ratio of distances covered in successive equal time intervals of $	au = 2 	ext{ s}$.
2. **Galileo's Law of Odd Numbers:** For a body falling from rest, the distances traversed during equal intervals of time stand to one another in the same ratio as the odd numbers beginning with unity .
   - Ratio = $1 : 3 : 5 : 7 : \dots$
3. **Verification via Calculation:**
   - Distance in 1st interval ($0-2	ext{ s}$): $h_1 = \frac{1}{2}g(2)^2 = 2g$.
   - Distance in 2nd interval ($2-4	ext{ s}$): Total distance in $4	ext{ s}$ ($8g$) minus distance in $2	ext{ s}$ ($2g$) = $6g$.
   - Distance in 3rd interval ($4-6	ext{ s}$): Total distance in $6	ext{ s}$ ($18g$) minus distance in $4	ext{ s}$ ($8g$) = $10g$.
   - Ratio: $2g : 6g : 10g = 1 : 3 : 5$.
4. **Check Validity:** The total distance fallen in $6	ext{ s}$ is approximately $180 	ext{ m}$ (taking $g=10 	ext{ m/s}^2$), which is less than the initial height of $200 	ext{ m}$, so the body is still in flight.

Step-by-Step Solution

Practice All PHYSICS Questions