Physics-
General
Easy

Question

A ball is dropped from a bridge at a height of 176.4 m over a river. After 2 s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously?

  1. 2.45 blank m s to the power of negative 1 end exponent  
  2. 49 blank m s to the power of negative 1 end exponent  
  3. 14.5 blank m s to the power of negative 1 end exponent  
  4. 24.5 blank m s to the power of negative 1 end exponent  

The correct answer is: 24.5 blank m s to the power of negative 1 end exponent


    For first ball
    fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent equals 176.4
    rightwards double arrow blank t equals blank square root of fraction numerator 176.4 blank cross times 2 over denominator 10 end fraction end root
    t equals 5.9 blank s
    For second ball, t equals 3.9 blank s
    u open parentheses 3.9 close parentheses plus fraction numerator 1 over denominator 2 end fraction g open parentheses 3.9 close parentheses to the power of 2 end exponent equals 176.4
    rightwards double arrow 3.9 u plus fraction numerator 10 over denominator 2 end fraction open parentheses 3.9 close parentheses to the power of 2 end exponent equals 176.4
    rightwards double arrow blank u equals 25.7 blank m s to the power of negative 1 end exponent
    This value is approximated to 24.5 blank m s to the power of negative 1 end exponent.

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