Question

# A stone is dropped from a height of 45m. What will be the distance travelled by it during the last second of its motion?(Take g=10msâˆ’2)

A
30 m
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B
25 m
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C
15 m
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D
5 m
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Solution

## The correct option is C 25 mWhen the stone is dropped, initial velocity = 0.Now, h=(1/2)gt2Distance travelled in tn sec can be written as hn=(1/2)gtn2Distance travelled in tn−1 sec can be written as hn−1=(1/2)g(tn−1)2So, hn−hn−1 = journey in nth second = (1/2)g(t2n−(tn−1)2) =(1/2)g(2tn−1)=g(tn−1/2)Now using h=(1/2)gt2 for h = 45m we get t = 3 secSo, in the 3rd second the stone will travel 10 X 2.5 m = 25 m.

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