Question

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 ?

• A. 4 : 5
• B. 7 : 9
• C. 16 : 25
• D. 1 : 1

Answer 1

Answer: (B)

7 : 9

Answer 2

Distance covered in $n^{th}$ second is given by
$s_n = u +\frac{a}{2} (2n-1)$
Given : u = 0, a = g
$\therefore \, \, \, s_4 = \frac{g}{2} (2 \times 4-1)=\frac{7g}{2}$
$s_5 = \frac{g}{2} (2 \times 5-1)=\frac{9g}{2} \, \, \, \, \, \therefore \frac{s_4}{s_5} =\frac{7}{9}$