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Question

After one second the velocity of a projectile makes an angle of 45 degree with the horizontal. After another one second it is travelling horizontally. The magnitude of its initial velocity and angle of projection are left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis

  1. 14.02 blank m s to the power of negative 1 end exponent comma tan to the power of negative 1 end exponent invisible function application left parenthesis 2 right parenthesis  
  2. 22.36 blank m s to the power of negative 1 end exponent comma tan to the power of negative 1 end exponent invisible function application left parenthesis 2 right parenthesis  
  3. 14.62 blank m s to the power of negative 1 end exponent comma blank 60 degree  
  4. 22.36 blank m s to the power of negative 1 end exponent comma blank 60 degree  

The correct answer is: 22.36 blank m s to the power of negative 1 end exponent comma tan to the power of negative 1 end exponent invisible function application left parenthesis 2 right parenthesis


    Time of flight of this particle, T equals 4 s. if u is its initial speed and theta is the angle of projection, then
    T equals 4 equals fraction numerator 2 u sin invisible function application theta over denominator g end fraction or u sin invisible function application theta equals 2 g(i)
    After 1 s, the velocity vector of particle makes an angle of 45 degree with horizontal, so
    v subscript x end subscript equals v subscript y end subscript blank i e comma u cos invisible function application theta equals open parentheses u sin invisible function application theta close parentheses minus g t
    or u cos invisible function application theta equals 2 g minus g blank left parenthesis therefore t equals 1 s right parenthesis
    or u cos invisible function application theta equals g(ii)
    Squaring and adding Eqs. (i) and (ii), we have
    u to the power of 2 end exponent equals 5 g to the power of 2 end exponent equals 5 open parentheses 10 close parentheses to the power of 2 end exponent equals 500
    or u equals square root of 500 equals 22.36 blank m s to the power of negative 1 end exponent
    Dividing Eq. (i) by Eq. (ii), we have
    tan invisible function application theta equals 2 blank o r blank theta equals tan to the power of negative 1 end exponent invisible function application open parentheses 2 close parentheses

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