Question

A ball falls free from rest. The ratio of distances traveled in first, second, third and fourth second is

• A. 1:1:1:2
• B. 1:2:3:4
• C. 1:1:1:3
• D. 1:3:5:7

Answer 1

The correct option is D

1:3:5:7

Step 1: Given:

1. A ball is free-falling from rest.
2. The initial velocity is zero.

Step 2: From the second equation of motion:

We know, distance traveled in nth second,

Where u is initial velocity, a is acceleration, t is time, and s is the distance traveled.

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ {s}_{n^{th}}=0+\frac{1}{2}a{(2n-1)}^2 \end{gathered}$$

= $\frac{1}{2}\left[g\right(2n-1\left)\right](gistheaccelerationduetogravity)$

Let$s_1,s_2,s_3,s_4$ be the distances traveled in the first, second, third, and fourth second respectively.

Step 3: Calculating$s_1,s_2,s_3,s_4$ :

$$\begin{gathered} s_1=\frac{1}{2}\left[g\right(2-1\left)\right]=\frac{1}{2}g \\ {s}_2=\frac{1}{2}\left[g\right(4-1\left)\right]=\frac{3}{2}g \\ {s}_3=\frac{1}{2}\left[g\right(6-1\left)\right]=\frac{5}{2}g \\ {s}_4=\frac{1}{2}\left[g\right(8-1\left)\right]=\frac{7}{2}g \end{gathered}$$

Therefore,

The Ratio is $s_1:s_2:s_3:s_4=1:3:5:7$

Hence, the correct option is (D).