A bullet is fired vertically upward with an initial velocity of $50 {ms}^{- 1}$ . If $g = 10 {ms}^{- 2}$ , what is the ratio of the distances travelled by the bullet during the first and last second of its upward motion ?

9 : 1

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9 : 2

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3 : 1

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9 : 4

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Solution

The correct option is A $9 : 1$
As we know that the time taken to reach the maximum height is
$t = \frac{u}{g} = \frac{50}{10} = 5 sec$
Also, displacement in $n^{t h}$ sec is given as
$S_{n^{t h}} = u + \frac{a}{2} [2 n - 1]$
As per the question we have
$\frac{S_{1}}{S_{5^{t h}}} = \frac{50 - \frac{10}{2} [2 (1) - 1]}{50 - \frac{10}{2} [2 (5) - 1]}$
$\Rightarrow \frac{S_{1}}{S_{5^{t h}}} = \frac{50 - 5}{50 - 45} = \frac{45}{5}$
$\Rightarrow \frac{S_{1}}{S_{5^{t h}}} = \frac{9}{1}$
Hence, the ratio is $9 : 1$

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