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A body is projected vertically upwards with a velocity 50 ms^-1. Ratio of distance traveled in first second of upward motion to first second of downward motion is

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1 : 7

5 : 3

9 : 1

3 : 5

answer is C.

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Detailed Solution

$\begin{array}{l} s_{up} = 50 \times 1 - \frac{1}{2} \times g {(1)}^{2} = 50 - 5 = 45 m; \\ s_{down} = \frac{1}{2} \times g {(1)}^{2} = \frac{g}{2} = = 5 m \\ ∴ \frac{S_{up}}{S_{down}} = \frac{45}{5} = 9 : 1 \end{array}$

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