A bullet is fired vertically upward with an initial velocity of 50ms−1. If g=10ms−2, what is the ratio of the distances travelled by the bullet during the first and last second of its upward motion ?
A
9:1
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B
9:2
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C
3:1
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D
9:4
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Solution
The correct option is A9:1 As we know that the time taken to reach the maximum height is t=ug=5010=5sec Also, displacement in nth sec is given as Snth=u+a2[2n−1] As per the question we have S1S5th=50−102[2(1)−1]50−102[2(5)−1] ⇒S1S5th=50−550−45=455 ⇒S1S5th=91 Hence, the ratio is 9:1