Physics-
General
Easy

Question

From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 blank m blank divided by s e c. The ratio of the distances, covered by it in the 3rd and 2nd seconds of the motion is (Take g equals 10 blank m divided by s to the power of 2 end exponent)

  1. 5 blank colon 7  
  2. 7 blank colon 5  
  3. 3 blank colon 6  
  4. 6 blank colon 3  

The correct answer is: 7 blank colon 5


    S subscript 3 to the power of r d end exponent end subscript equals 10 plus fraction numerator 10 over denominator 2 end fraction open parentheses 2 cross times 3 minus 1 close parentheses equals 35 blank m
    S subscript 2 to the power of n d end exponent end subscript equals 10 plus fraction numerator 10 over denominator 2 end fraction open parentheses 2 cross times 2 minus 1 close parentheses equals 25 blank m rightwards double arrow fraction numerator S subscript 3 to the power of r d end exponent end subscript over denominator S subscript 2 to the power of n d end exponent end subscript end fraction equals fraction numerator 7 over denominator 5 end fraction

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