Question

Q.1 Find the ratio of distance travelling by a freely falling body in first,second & third second of its fall . [ NCERT CLASS 11]

Answer 1

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we know that

s = ut + (1/2)at2

 

in the first case

now, here u = 0

t1 = 1 s

a = 9.8 m/s2


so,

s1 = (1/2)x9.8x12

or

the distance travelled after 1st second will be

s1 = 4.9 m

here, the velocity after 1st second will be 

v1 = at1

or

v1 = 9.8x1 = 9.8 m/s

 
now, in the second case

s2 = v1t2+ (1/2)at2^2

here

v1 = 9.8 m/s

a = 9.8 m/s2

t2 = 1 s

thus,

s2 = 9.8x1 + (1/2)x9.8x12

or

the distance travelled after 2nd second will be

s2 = 14.7 m

the velocity after the 2nd second will be

v2 = v1 + at2

or

v2 = 9.8 + 9.8x1

thus,

v2 = 19.6 m/s

 
similarly,

the distance travelled after 3rd second will be

s3 = v2t^3 + (1/2)at3^2

here

v1 = 19.6 m/s

a = 9.8 m/s2

t3 = 1 s

or

the distance travelled after 3rd second will be

s3 = 19.6 m


thus, the ratio will be

s1:s2:s3 = 4.9:14.7:19.6 = 1:3:4

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Answer 2

ANSWER:-

➡️The ratio is 1:3:5:7... and This is known as law of odd numbers proved by The great Galieleo.⬅️

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