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.