Question

A body falling from rest describes distances s1,s2,s3 in the first, second and third seconds of its fall then the ratio s1:s2:s3 is

Answer 1

Then the ratio of S1 : S2 : S3 is : (2) 1 : 3 : 5

The distance travelled are 5m, 15m and 25m.
(Taking g=10 m/s²)

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Answer 2

Answer:

The answer is 1:3:5

Explanation:

Using $s=ut+\frac{1}{2}at^2$,

Distance traveled by the body in the first second= $0*1+\frac{1}{2}*10*1^2$

                                                                                 =5 meters=$S_1$

Distance traveled by the body in the second second= distance traveled by the body in two seconds- distance traveled by the body in one second

=$0*2+\frac{1}{2}*10*2^2-5$

=20-5

=15 meters=$S_2$

Distance traveled by the body in the third second= distance traveled by the body in three seconds- distance traveled by the body in two seconds

=$0*3+\frac{1}{2}*10*3^2-20$

=45-20

=25 meters=$S_3$

So, answer is...

$S_1:S_2:S_3\\=5:15:25\\=(5*1):(5*3):(5*5)=1:3:5$