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Question: A particle starting from rest falls from a certain height. Assuming that the acceleration due to gra...

A particle starting from rest falls from a certain height. Assuming that the acceleration due to gravity remain the same throughout the motion, its displacements in three successive half second intervals are S1,S2S_1, S_2 and S3S_3 then

A

S1:S2:S3=1:5:9S_1:S_2:S_3 = 1:5:9

B

S1:S2:S3=1:3:5S_1:S_2:S_3 = 1:3:5

C

S1:S2:S3=9:2:3S_1:S_2:S_3 = 9:2:3

D

S1:S2:S3=1:1:1S_1:S_2:S_3 = 1:1:1

Answer

1:3:5

Explanation

Solution

Let the particle start from rest at t=0t=0. The initial velocity is u=0u=0.
The acceleration due to gravity is constant, let it be gg downwards.
The displacement of the particle at time tt is given by the equation of motion:
s=ut+12at2s = ut + \frac{1}{2}at^2
Since u=0u=0 and a=ga=g, the displacement from the starting point at time tt is:
s(t)=12gt2s(t) = \frac{1}{2}gt^2

We are given three successive half-second intervals. Let these intervals be:
Interval 1: from t=0t=0 to t=0.5t=0.5 s
Interval 2: from t=0.5t=0.5 s to t=1.0t=1.0 s
Interval 3: from t=1.0t=1.0 s to t=1.5t=1.5 s

The displacement in the first half-second interval (S1S_1) is the displacement from t=0t=0 to t=0.5t=0.5 s.
S1=s(0.5)−s(0)=12g(0.5)2−12g(0)2=12g(0.25)=g8S_1 = s(0.5) - s(0) = \frac{1}{2}g(0.5)^2 - \frac{1}{2}g(0)^2 = \frac{1}{2}g(0.25) = \frac{g}{8}

The displacement in the second half-second interval (S2S_2) is the displacement from t=0.5t=0.5 s to t=1.0t=1.0 s.
S2=s(1.0)−s(0.5)=12g(1.0)2−12g(0.5)2=g2−g8=3g8S_2 = s(1.0) - s(0.5) = \frac{1}{2}g(1.0)^2 - \frac{1}{2}g(0.5)^2 = \frac{g}{2} - \frac{g}{8} = \frac{3g}{8}

The displacement in the third half-second interval (S3S_3) is the displacement from t=1.0t=1.0 s to t=1.5t=1.5 s.
S3=s(1.5)−s(1.0)=12g(1.5)2−12g(1.0)2=2.25g2−g2=1.25g2=5g8S_3 = s(1.5) - s(1.0) = \frac{1}{2}g(1.5)^2 - \frac{1}{2}g(1.0)^2 = \frac{2.25g}{2} - \frac{g}{2} = \frac{1.25g}{2} = \frac{5g}{8}

Now we find the ratio of the displacements S1:S2:S3S_1:S_2:S_3.
S1:S2:S3=g8:3g8:5g8S_1:S_2:S_3 = \frac{g}{8} : \frac{3g}{8} : \frac{5g}{8}
Multiplying the ratio by 8g\frac{8}{g} (assuming g≠0g \neq 0), we get:
S1:S2:S3=1:3:5S_1:S_2:S_3 = 1:3:5

This result aligns with Galileo's law of odd numbers, which states that for a particle starting from rest under constant acceleration, the distances covered in successive equal time intervals are in the ratio of odd numbers 1:3:5:7:…1:3:5:7:\dots.