Question

The distances traveled by a body falling freely from rest in the first, third and fifth second are in the ratio
(A) 1:9:25
(B) 1:3:5
(C) 1:4:9
(D) 1:5:9​

Answer 1

Answer:

B) 1:3:5

Explanation:

we know that a freely falling body has its distance doubled but a ratio can be made in simplest form so this is the anwer

Answer 2

Initial velocity of the body u=0

Distance covered in t^ th second St = u+

$\frac{1}{2} g (2t - 1)$

$∴  St=0+ \frac{1}{2} g(2t−1)= \frac{1}{2} g(2t−1)$

Distance travelled in first second i.e. t=1,

$S1= \frac{1}{2} g(2×1−1)= \frac{ 1}{2} g$

Distance travelled in 2nd second i.e. t=2,

$S2= \frac{1}{2} g(2×2−1)= \frac{3}{2} g$

Distance travelled in third second i.e. t=3,

$S3= \frac{1}{2} g(2×3−1)= \frac{5}{2} g</p><p> \\ </p><p>⟹  S1:S2:S3=1:3:5 \\ </p><p></p><p>$