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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=10ms2)

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 m
When the stone is dropped, initial velocity = 0.
Now, h=(1/2)gt2
Distance travelled in tn sec can be written as hn=(1/2)gtn2
Distance travelled in tn1 sec can be written as hn1=(1/2)g(tn1)2
So, hnhn1 = journey in nth second = (1/2)g(t2n(tn1)2)
=(1/2)g(2tn1)=g(tn1/2)
Now using h=(1/2)gt2 for h = 45m we get t = 3 sec
So, in the 3rd second the stone will travel 10 X 2.5 m = 25 m.

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