From top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s . ratio of

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2.Motion in Straight Line

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From the top of a tower, a particle is thrown vertically downwards with a velocity of $10\; m/s$. The ratio of the distances, covered by it in the $3^{rd}$ and $2^{nd}$ seconds of the motion is (Take $g = 10\,m/{s^2}$)

A

$5:7$

B

$7:5$

C

$3:6$

D

$6:3$

(AIIMS-2000)

Solution

(b) ${S_{{3^{rd}}}} = 10 + \frac{{10}}{2}(2 \times 3 – 1) = 35\;m$

${S_{{2^{nd}}}} = 10 + \frac{{10}}{2}(2 \times 2 – 1) = 25\,m$

$\frac{{{S_{{3^{rd}}}}}}{{{S_{{2^{nd}}}}}} = \frac{7}{5}$

Standard 11

Physics

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