Question

A stone falls freely under gravity. It covers distances $h_1, h_2$ and $h_3$ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between $h_1, h_2$ and $h_3$ is :-

• A. $h_1 = h_2 =h_3$
• B. $h_1 = 2 h_2 = 3h_3$
• C. $h_1 = \frac{ h_2}{3} = \frac{h_3}{5}$
• D. $h_2 = 3h_1 $ and $h_3 = 3 h_2$

Answer 1

The Correct Option is C

To solve this problem, we need to determine how much distance a freely falling stone covers in successive time intervals of equal duration. The key here is the understanding of motion under gravity.

The distances covered by the stone in the $n$th time interval of equal duration can be calculated using the equations of motion. For a freely falling object, the displacement $s$ covered in time $t$ is given by:

s = ut + \frac{1}{2}gt^2

where $u$ is the initial velocity (which is zero for free fall), $g$ is the acceleration due to gravity, and $t$ is the time.

Now, using this formula for each interval:

1. First 5 seconds: At $t = 5\, \text{seconds}$
   • The stone starts from rest, hence $u=0$.
   • h_1 = \frac{1}{2} \cdot g \cdot (5)^2 = \frac{1}{2} \cdot g \cdot 25 = 12.5g
2. Next 5 seconds: From $t = 5$ to $t = 10\, \text{seconds}$
   • Total distance at $t = 10$ seconds: h = \frac{1}{2} \cdot g \cdot (10)^2 = 50g
   • h_2 = 50g - 12.5g = 37.5g
3. Next 5 seconds: From $t = 10$ to $t = 15\, \text{seconds}$
   • Total distance at $t = 15$ seconds: h = \frac{1}{2} \cdot g \cdot (15)^2 = 112.5g
   • h_3 = 112.5g - 50g = 62.5g

Thus, we have found the distances:

- h_1 = 12.5g

- h_2 = 37.5g

- h_3 = 62.5g

Now, let's identify the relation between h_1, h_2, and h_3 using these values:

- h_1 = \frac{h_2}{3} = \frac{h_3}{5}

This demonstrates that the correct relationship is $h_1 = \frac{ h_2}{3} = \frac{h_3}{5}$, which corresponds to the correct answer.