Question

An object is dropped from a certain point $A$ at a height 'h' from the ground. During its journey straight downwards, the object passes points $B$ and $C$ such that the ratio of time taken $t_1$ to cover $A B$ and $t_2$ to cover $B C$ is $1 : (\sqrt{2}-1)$. What is the ratio of distances $A B : B C$?

• A. $\sqrt{2} : 1$
• B. $(\sqrt{2}-1) : 1$
• C. $1 : 1$
• D. $1 : \sqrt{2}$

Hint:

For any object dropping from rest, the times taken to cover consecutive equal distances are always in the specific ratio: $1 : (\sqrt{2}-1) : (\sqrt{3}-\sqrt{2}) : \dots$ Since the provided times match this exact pattern, the distances must be equal!

Answer 1

The Correct Option is C

Concept: For an object performing free-fall motion under constant gravity from rest ($u = 0$), the total distance traversed ($S$) in total time $t$ is described by the kinematic equation: \[ S = \frac{1}{2}gt^2 \implies S \propto t^2 \]

Step 1: Set up expressions for distances using elapsed times.
Let the time taken to travel distance $AB$ be $t_1 = t$. The question states that the time to travel segment $BC$ is $t_2 = (\sqrt{2}-1)t$. Therefore, the total time to travel from the starting point $A$ all the way to $C$ is: \[ t_{\text{total}} = t_1 + t_2 = t + (\sqrt{2}-1)t = \sqrt{2}t \]

Step 2: Formulate the distance ratio.
Using the proportionality law for uniform acceleration: \[ \text{Distance } AB = \frac{1}{2}gt_1^2 = \frac{1}{2}gt^2 \] \[ \text{Total Distance } AC = \frac{1}{2}gt_{\text{total}}^2 = \frac{1}{2}g(\sqrt{2}t)^2 = 2 \left(\frac{1}{2}gt^2\right) \] The remaining intermediate distance section $BC$ is: \[ BC = AC - AB = 2 \left(\frac{1}{2}gt^2\right) - \frac{1}{2}gt^2 = \frac{1}{2}gt^2 \]

Step 3: Compute the final ratio.
Comparing the two segments: \[ \frac{AB}{BC} = \frac{\frac{1}{2}gt^2}{\frac{1}{2}gt^2} = \frac{1}{1} \implies 1:1 \]