Motion in a Straight Line - Result Question 49
51. What will be the ratio of the distances moved by a freely falling body from rest in 4th and 5th seconds of journey?
[1989]
(a) $4: 5$
(b) $7: 9$
(c) $16: 25$
(d) $1: 1$
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Answer:
Correct Answer: 51. (b)
Solution:
- (b) $\frac{x(4)}{x(5)}=\frac{\frac{g}{2}(2 \times 4-1)}{\frac{g}{2}(2 \times 5-1)}=\frac{7}{9}$
$[\because S _{n^{\text{th }}}=u+\frac{a}{2}(2 n-1).$ and $.u=0, a=g]$
In the absence of air resistance, all bodies irrespective of the size, weight or composition fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small altitude $(h«R)$ is called free fall.
Distance travelled in nth second $h_n=\frac{g}{2}(2 n-1)$
