What will be the ratio of the distance moved by a freely falling body from rest in 4 textth and 5 textth seconds of journey ?

Question

What will be the ratio of the distance moved by a freely falling body from rest in 4 textth and 5 textth seconds of journey ? (A) 4: 5 (B) 7: 9 (C) 16: 25 (D) 1: 1

Tardigrade Admin · Accepted Answer

Distance covered in nt h second is given by sn=u+(a/2)(2 n-1) Given: u = 0 , a = g ∴ s4=(g/2)(2 × 4-1)=(7 g/2) s5=(g/2)(2 × 5-1)=(9 g/2) ∴ (s4/s5)=(7/9)

Q. What will be the ratio of the distance moved by a freely falling body from rest in $4^{th}$ and $5^{th}$ seconds of journey ?

4 : 5

7 : 9

16 : 25

1 : 1

Distance covered in $n^{t h}$ second is given by
$s_{n} = u + \frac{a}{2} (2 n - 1)$
Given : $u = 0, a = g$
$∴ s_{4} = \frac{g}{2} (2 \times 4 - 1) = \frac{7 g}{2}$
$s_{5} = \frac{g}{2} (2 \times 5 - 1) = \frac{9 g}{2}$
$∴ \frac{s _{4}}{s _{5}} = \frac{7}{9}$

Q. What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey ?