Physics-
General
Easy

Question

A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

  1. fraction numerator 4 v to the power of 2 end exponent over denominator 5 g end fraction  
  2. fraction numerator 4 g over denominator 5 v to the power of 2 end exponent end fraction  
  3. fraction numerator v to the power of 2 end exponent over denominator g end fraction  
  4. fraction numerator 4 v to the power of 2 end exponent over denominator square root of 5 g end fraction  

The correct answer is: fraction numerator 4 v to the power of 2 end exponent over denominator 5 g end fraction


    R equals 2 H Given
    We know R equals 4 H cot invisible function application theta rightwards double arrow cot invisible function application theta equals fraction numerator 1 over denominator 2 end fraction
    From triangle we can say that sin invisible function application theta equals fraction numerator 2 over denominator square root of 5 end fraction comma cos invisible function application theta equals fraction numerator 1 over denominator square root of 5 end fraction
    thereforeRange of projectile R equals fraction numerator 2 v to the power of 2 end exponent sin invisible function application theta cos invisible function application theta over denominator g end fraction
    equals fraction numerator 2 v to the power of 2 end exponent over denominator g end fraction cross times fraction numerator 2 over denominator square root of 5 end fraction cross times fraction numerator 1 over denominator square root of 5 end fraction equals fraction numerator 4 v to the power of 2 end exponent over denominator 5 g end fraction

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