Question

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

Answer 1

The ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey is 7 : 9.

We have to find the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey.

We know, distance covered by a body in nth second in an accelerating motion is given by,

$S_n=u+\frac{1}{2}a(2n-1)$

Where u is initial velocity  in m/s , a is acceleration in m/s² and n is nth second.

For free falling body,

Initial velocity, u = 0

acceleration of body, a = g = 9.8m/s²

∴ Distance covered by  body in 4th second, S₄ = 0 + 1/2 × 9.8 × (2 × 4 - 1) = 4.9 × 7

And distance covered by body in 5th second, S₅ = 0 + 1/2 × 9.8 × (2 × 5 - 1) = 4.9 × 9

∴ The ratio of the distance covered in 4th and 5th second is S₄ : S₅ = 4.9 × 7 : 4.9 × 9 = 7 : 9

Therefore the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey is 7 : 9.

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Answer 2

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