• # question_answer 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 3rd and 2nd seconds of the motion is (Take$g=10\text{ }m/{{s}^{2}}$) A)  5:7         B)  7:5C)  3:6         D)  6:3

Correct Answer: B

Solution :

${{S}_{3rd}}=10+\frac{10}{2}(2\times 3-1)=35\,m$ ${{S}_{2nd}}=10+\frac{10}{2}(2\times 2-1)=25\,m\Rightarrow \frac{{{S}_{3rd}}}{{{S}_{2nd}}}=\frac{7}{5}$

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