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Two projectiles of same mass have their maximumkinetic energies in ratio 4 1 and ratio of theirmaximum heights is also 4 I then what is theratio of their ranges?a) 2(d) 16: 1 |
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Answer» if we have a particle with velocity v, then kinetic energy, K = 1/2 m v^2 max height attained by this ,H = v^2 / 2 g but in this 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 (a) option. |
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