(B) The distance covered in the $n^{th}$ second is given by the formula: $S_n = u + \frac{g}{2}(2n - 1)$.Given: Initial velocity $u = 10\; m/s$,accele

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(B) The distance covered in the $n^{th}$ second is given by the formula: $S_n = u + \frac{g}{2}(2n - 1)$.Given: Initial velocity $u = 10\; m/s$,acceleration $g = 10\; m/s^2$.For the $3^{rd}$ second $(n=3)$:$S_3 = 10 + \frac{10}{2}(2 \times 3 - 1) = 10 + 5(5) = 10 + 25 = 35\; m$.For the $2^{nd}$ second $(n=2)$:$S_2 = 10 + \frac{10}{2}(2 \times 2 - 1) = 10 + 5(3) = 10 + 15 = 25\; m$.The ratio of the distances is $\frac{S_3}{S_2} = \frac{35}{25} = \frac{7}{5}$.

Class 11 Physics Motion in Straight Line Motion Under Gravity

Medium

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 $g = 10\; m/s^2$).

AIIMS 2000 All AIIMS

A
$5:7$
B
$7:5$
C
$3:6$
D
$6:3$

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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 $g = 10\; m/s^2$).

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