Question

2. The ratio of distances travelled by a body starting from rest with constant acceleration in 9th and 8th second is



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Answer 1

Answer:

ratio of distance traveled in 9th sec. to 8th sec =

$\frac{17}{15}$

Explanation:

Distance traveled by a free a free falling body in nth second =

distance traveled in n secs−distance traveled in n−1 secs.

$= 0(n) + \frac{1}{2} g {n}^{2} - 0(n - 1) + \frac{1}{2} g {(n - 1)}^{2}$

$= \frac{g}{2} (2n - 1)$

ratio of distance traveled in 9th sec. to 8th sec =

$\frac{2(9) - 1}{2(8) - 1} = \frac{17}{15}$

Answer 2

Answer:

24 is correct answer

Explanation:

i hope this help