Question

$\tt\large\underline{Question:-}$


A stone is falling freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first 3 seconds of its motion. How long does the stone remain in air?
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Answer 1

Answer:

5 s, 122.5m

Explanation:

Sn= distance coveredin first 3s 

Sn=u×t+21×10×32

Freely ball  so, u=0

Sn=21×10×=9=45m

Sn=4+29(2t−1)=45

t=5s in air

h=21gt2=4.9×25=122.5m

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Answer 2

$\tt\large\underline{Question:-}$

A stone is falling freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first 3 seconds of its motion. How long does the stone remain in air?

$\tt\large\underline{Answer:-}$

$s \: = \: ut \: + \: {0.5gt}^{2} \: \: \: \: \: \: - > \: 1$

In first 3 seconds, let it covers distance "d"m.

$d \: = \: 0.5g {(3)}^{2} = \: 4.5g$

The distance is same as distance traveled in last second.

So, according to formula,

Sn = u + 0.5a ( 2n - 1 )

4.5g = 0.5g ( 2n - 1 )

Here, u = 0 and n is number of seconds for which stone is in air.

ஃ n = 5sec

Substituting in eqn 1,

S = 122.5m