Question

The ratio of the displacements of a freely falling body during second and fifth seconds of its motion is

• A. 1 : 1
• B. 2 : 5
• C. 4 : 25
• D. 1 : 3

Hint:

For motion under gravity from rest, the displacement in the \(n^th}\) second is given by: \[ s_n = \frac{1}{2}g(2n - 1) \] This formula helps quickly determine displacement during specific seconds.

Answer 1

The Correct Option is D

Step 1: Use the formula for displacement in the \(n^{\text{th}}\) second of free fall.
Displacement in the \(n^{\text{th}}\) second is given by:
\[ s_n = u + \frac{1}{2}g(2n - 1) \] For free fall from rest, \( u = 0 \), so:
\[ s_n = \frac{1}{2}g(2n - 1) \] Step 2: Calculate displacement in the 2^nd and 5^th seconds.
\[ s_2 = \frac{1}{2}g(2 \cdot 2 - 1) = \frac{1}{2}g(3) = \frac{3g}{2} \] \[ s_5 = \frac{1}{2}g(2 \cdot 5 - 1) = \frac{1}{2}g(9) = \frac{9g}{2} \] Step 3: Take the ratio of displacements.
\[ \frac{s_2}{s_5} = \frac{\frac{3g}{2}}{\frac{9g}{2}} = \frac{3}{9} = \frac{1}{3} \] Step 4: Select the correct option.
The ratio of displacements is \( 1 : 3 \), which corresponds to option (4).