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Step-by-Step Solution
Step 1: Identify the physical situation
A stone is falling freely under gravity, which means it is undergoing uniformly accelerated motion downwards with acceleration $g$ (no initial velocity). We are told it covers distances $h_1$, $h_2$, and $h_3$ in successive 5-second intervals:
• $h_1$ in the first 5 seconds (from $t=0$ to $t=5$),
• $h_2$ in the next 5 seconds (from $t=5$ to $t=10$),
• $h_3$ in the following 5 seconds (from $t=10$ to $t=15$).
Step 2: Write down the formula for distance under constant acceleration
For an object starting from rest and moving under constant acceleration $g$, the distance $s$ covered in time $t$ is given by
$$ s = \frac{1}{2} g t^2. $$
Step 3: Calculate $h_1$ (distance in the first 5 seconds)
The distance covered from $t=0$ to $t=5$ is
$$ h_1 = \frac{1}{2} g (5)^2 = \frac{1}{2} g \times 25 = 12.5\,g. $$
Step 4: Calculate $h_2$ (distance in the next 5 seconds, i.e., from $t=5$ to $t=10$)
The total distance covered from $t=0$ to $t=10$ is
$$ s_{0\rightarrow 10} = \frac{1}{2} g (10)^2 = 50\,g. $$
Hence,
$$ h_2 = s_{0\rightarrow 10} - s_{0\rightarrow 5} = 50\,g - 12.5\,g = 37.5\,g. $$
Step 5: Calculate $h_3$ (distance in the next 5 seconds, i.e., from $t=10$ to $t=15$)
The total distance covered from $t=0$ to $t=15$ is
$$ s_{0\rightarrow 15} = \frac{1}{2} g (15)^2 = \frac{1}{2} g \times 225 = 112.5\,g. $$
Therefore,
$$ h_3 = s_{0\rightarrow 15} - s_{0\rightarrow 10} = 112.5\,g - 50\,g = 62.5\,g. $$
Step 6: Relate $h_1$, $h_2$, and $h_3$
From the above results:
$ h_1 = 12.5\,g $
$ h_2 = 37.5\,g $
$ h_3 = 62.5\,g $
We observe that
$ \frac{h_2}{3} = \frac{37.5\,g}{3} = 12.5\,g = h_1, $
$ \frac{h_3}{5} = \frac{62.5\,g}{5} = 12.5\,g = h_1. $
Hence, the correct relationship is
$$ h_1 = \frac{h_2}{3} = \frac{h_3}{5}. $$
Step 7: Conclusion
The distances covered in three consecutive 5-second intervals by a freely falling stone have the ratio
$$ h_1 : h_2 : h_3 = 1 : 3 : 5,$$
which matches Option 4 in the question.
