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Question

A body is projected up with a velocity 96m/s . The ratio of distance covered by the body in last two seconds of its motion and last two seconds of its ascent.

A
1:7
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B
7:9
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C
9:1
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D
3:5
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Solution

The correct option is A 1:7

Velocity of a particle during the last second of its ascent will be 9.8m/s

And the velocity of the same particle in the second before that will be 19.6m/s

So the distance covered in both the seconds will be

9.8+19.6m=29.4m

Velocity of a particle during the last second of its descent will be 96m/s

And the velocity of the same particle in the second before that will be 87.2m/s

So the distance covered in both the seconds will be

96+87.2m=183.2m

The ratio of the distance covered during ascent and descent will be,

=29.4183.2

17


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