Physics-
General
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Question

An airplane, diving at an angle of 53.0 degree with the vertical releases a projectile at an altitude of 730 m. The projectile hits the ground 5.00 s after being released. What is the speed of the aircraft?

  1. 282 blank m s to the power of negative 1 end exponent  
  2. 202 blank m s to the power of negative 1 end exponent  
  3. 182 blank m s to the power of negative 1 end exponent  
  4. 102 blank m s to the power of negative 1 end exponent  

The correct answer is: 202 blank m s to the power of negative 1 end exponent


    Since the projectile is released its initial velocity is the same as the velocity of the plane at the time of release
    Take the origin at the point of release
    Let x and y left parenthesis equals negative 730 m right parenthesis be the coordinates of the point on the ground where the projectile hits and let t be the time when it hits. Then
    y equals negative v subscript 0 end subscript t cos invisible function application theta minus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
    where theta equals 53.0 degree
    This equation gives
    v subscript 0 end subscript equals negative fraction numerator y plus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent over denominator t cos invisible function application theta end fraction
    equals negative fraction numerator negative 730 plus fraction numerator 1 over denominator 2 end fraction open parentheses 9.8 close parentheses open parentheses 5 close parentheses to the power of 2 end exponent over denominator 5 cos invisible function application 53 degree end fraction equals 202 blank m s to the power of negative 1 end exponent

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