hi,
    if we have a partical with velocity v then 
                            kinetic energy, K = 1/2 m v^2 
                           max height attained by this ,H = v^2 / 2 g
     bt inthis case velocity must be vertical , otherwise we have to use its vertical component , v sin x  instead of it in the H expression
                    
                       so , we can easily get , k = mg H ,



problem solution,
          
         let the velocities of two particals be v1, v2
                   angle with horizontal be x1 ,x2
                     kinetic energies  k1, k2
                      max. height     H1, H2 
                  & range r1 ,r2 

   k1/k2 = 4 /1 , gives
    v1 /v2 = 2/1
we know  that,
  H1 =  (v1sinx1 )^2 /2g
   H2 =  (v2sinx2 )^2 /2g

  so H1 / H2 = 4/ 1
gives  v1sinx1 / v2sinx2  =  2/ 1
 from eq (1),.............
  
 we get sinx1 = sinx2 
 so, x1 = x2


range of first r1 =  2v1 sinx cosx / g
          second, r2 = 2v2 sinx cosx / g


so, r1/ r2 = v1/ v2 = 2/1

ans is d option.



     
						

							
The kinetic energies are minimum at the maximum height where velocity is ucosx
Hence 1/2 mu^2cos^2x1= 4*1/2 mv^2cos^2x2
or, ucosx1 = 2vcosx2........(1)
Maximum height is given by H = u^2sin^x/2g
Hence u^2sin^2x1/2g = 4*v^2sin^2x2/2g
or, usinx1 = 2vsinx2.......(2)
(2)/(1) gives tanx1=tanx2 or x1=x2
Using in (1) we get ucosx1 = 2vcosx1
hence u=2v
Therefore R1/R2 = {u^2sin2x1/g}/{v^2sin2x2/g} = (2v)^2sin2x1/v^2sin2x2  = 4
Hence option B is correct.
						

							
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hope it might help you
 
 
 
 
 
 
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