The ratio of the distance travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second [2022]

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Solution

Q.
The ratio of the distance travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second [2022]

1 1 : 2 : 3 : 4

2 1 : 4 : 9 : 16

3 1 : 3 : 5 : 7

4 1 : 1 : 1 : 1

Ans.
(3)

Distance travelled by a body during free fall is given by

$S = u t + \frac{1}{2} a t^{2}$

Here, $u = 0$ and $a = g$

$∴$ $S \propto t^{2}$

Here for 1st second,

   $S_{1} = K t^{2} = K$

For 2nd second,

   $S_{2} = K {(2)}^{2} = 4 K$

For 3rd second,

   $S_{3} = K {(3)}^{2} = 9 K$

For 4th second,

   $S_{4} = K {(4)}^{2} = 16 K$

Distance covered in 2nd seconds, $4 K - K = 3 K$

Distance covered in 3rd second, $9 K - 4 K = 5 K$

Distance covered in 4th second, $16 K - 9 K = 7 K$

$∴$   Ratio of distances travelled by the freely falling body will be $1 : 3 : 5 : 7$

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