Question

a body falling from rest describes distance S1 s2 and s3 is a first second and third seconds of its fall in the ratio 1 ratio 2 is to 3 is

Answer 1

The ratio is 1:3:5.


[Read this if you want to see the formulas.

Using the two equations for constant acceleration:

1. s=ut+12at2s=ut+12at2
2. v=u+atv=u+at

In the first second, u=0u=0, t=1t=1 s, a=−ga=−g, so s=−g/2s=−g/2 and v=−gv=−g.

In the second second, u=−gu=−g, t=1t=1 s, a=−ga=−g, so s=−g−g/2=−3g/2s=−g−g/2=−3g/2 and v=−g−g=−2gv=−g−g=−2g.

In the third second, u=−2gu=−2g, t=1t=1 s, a=−ga=−g, so s=−2g−g/2=−5g/2s=−2g−g/2=−5g/2 and v=−2g−g=−3gv=−2g−g=−3g.

Answer 2

Answer:

The answer is 1 : 3 : 5

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Explanation:

Let u be the initial velocity of the body.

Then S= ut + $\frac{1}{2\\}$ $at^{2}$

For distance S1,

S1=$\frac{1}{2}$×g×$1^{2}$=$g^{2}$    .....(1)

Now,S1+S2=12×g×$2^{2}$ = 2g     .....(2)

and

S1+S2+S3=12×g×$3^{2}$ = 4.5g     .....(3)

Subtracting (1) from (2), we haveS1+S2−S1=2g−g2⇒S2=1.5g

Substracting (2) from (3), we haveS3=2.5g

Therefore,

S1 : S2 : S3 = 12g : 1.5g : 2.5g = 1 : 3 : 5

Hence, the correct answer is 1 : 3 : 5.

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