From the top of a tower, a particle is thrown vertically downward… - Vidyadip

MCQ

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
$5:7$
✓
$7:5$
C
$3:6$
D
$6:3$

✓

Answer

Correct option: B.

$7:5$

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

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

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

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