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Easy

Question

A particle A is projected from the ground with an initial velocity of 10 blank m s to the power of negative 1 end exponentat an angle of 60degree with horizontal. From what height h should an another particle B be projected horizontally with velocity 5 m s to the power of negative 1 end exponent so that both the particles collide in ground at point C if both are projected simultaneously? left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis

  1. 10 m    
  2. 30 m    
  3. 15 m    
  4. 25 m    

The correct answer is: 15 m


    Horizontal component of velocity of A is 10 c o s blank 6 0 degree or 5 blank m s to the power of negative 1 end exponent which is equal to the velocity of B in horizontal direction. They will collide at C if time of flight of the particles are equal or t subscript A end subscript equals t subscript B end subscript
    fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses
    or h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction
    equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m

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