Kinematics Question 230

Question: 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=10m/s^{2} $ ) [AIIMS 2000; CBSE PMT 2002]

Options:

A) 5 : 7

B) 7 : 5

C) 3 : 6

D) 6 : 3

Show Answer

Answer:

Correct Answer: B

Solution:

$ {S _{3^{rd}}}=10+\frac{10}{2}(2\times 3-1)=35\ m $

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

therefore $ \frac{{S _{3^{rd}}}}{{S _{2^{nd}}}}=\frac{7}{5} $



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