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/s2 ) [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:

S3rd=10+102(2×31)=35 m

S2nd=10+102(2×21)=25m

therefore S3rdS2nd=75



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