Physics-
General
Easy

Question

Two bodies are thrown simultaneously from a tower with same initial velocity v subscript 0 end subscript: one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is

  1. 2 v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent  
  2. 2 v subscript 0 end subscript t  
  3. v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent  
  4. v subscript 0 end subscript t  

The correct answer is: 2 v subscript 0 end subscript t


    For vertically upward motion, h subscript 1 end subscript equals v subscript 0 end subscript t minus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent and for vertically downward motion, h subscript 2 end subscript equals v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction g blank t to the power of 2 end exponent
    therefore T o t a l blank d i s t a n c e blank c o v e r e d blank i n blank t sec invisible function application h equals h subscript 1 end subscript plus h subscript 2 end subscript equals 2 v subscript 0 end subscript t

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