Question

the distance covered by a stone dropped from the top of the building in the last second of motion is 0.36 times the total distance travelled in the height of building (
g=m/s^2) ​

Answer 1

Answer:

Distance covered in 3 sec=

2

1

gt

2

=

2

1

×10×3

2

=45m

At the end of 'n' second its velocity u=gn

At the end of 'n+1' second its velocity v=g(n+1)

Distance traveled in the (n+1)th second=

2

(2n+1)g

=

2

(20n+10)

=10n+5

10n+5=45⇒(n+1)=5

Distance covered in 5 second =

2

1

gt

2

=

2

1

×10×5

2

=125m

Explanation:

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Answer 2

Explanation:

125m.

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