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en projected at an angle of projection for maximum range. Its K.E. at theA body has kinetic energy E whhighest point of its path will be:219.(A) E2(D) zero |
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Answer» For maximum range the value of angle is 45° , assume initial velocity as V. so, E = kV².. k is some constant. and at the maximum height, only horizontal part of velocity is left that is Vcos∅ = Vcos45 =V/√2. now, K.E at top is E' = k(V/√2)² = kV²/2. but kV² = E so E' = E/2. option C |
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