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  # Solve this
  
  Asked by venkat 25th December 2017, 8:56 PM
  Since Q is projected after 1 second, P will reach the maximum height and meet Q when it is returning back.
  
  Let maximum height reached by P is H.
  
  H = (u×u)  /(2×g), where u is initial projection velocity 30 m/s
  
  H = (30×30) / (2*10) = 45 m.
  
  after reaching the maximumheight let P travels back for t seconds and meet Q.
  
  Let the distance travelled by P in that t seconds is h and it is given by
  
  h = (1/2)g×t2 = (1/2)×10×t2 = 5×t2 .........................(1)
  
  Then distance travelled by Q when it meets P is 45-h and the time taken is (2+t)s. This distance is given by
  
  45-h = 30×(2+t) - (1/2)×10×(2+t)2
  
  after substituting for h from eqn.(1), we will get
  
  45 - 5×t2 = 30×(2+t)-5×(2+t)2
  
  solving the above eqn for t, we get t = 0.5 s.
  
  Hence P travelled 3.5 s before meetin Q
  Answered by Expert 27th December 2017, 3:13 PM
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