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A body is thrown upward at intial velocity of 7 ms^(-1) , then at which height it's kinetic energy becomes half ? |
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Answer» Solution :Suppose at height h its KINETIC becomeshalveto its initial kinetic energy . `K = 1/2 (K_(0))/2 ` `1/2 mv^(2) = 1/2 (1/2mv_(0)^(2))` ` :. V^(2) = 1/2 v_(0)^(2) = 2gh""…(1)` but equation of a body thrown in upward direction . `v_(2)^(2) -v_(0)^(3) =2gh` `1/2v_(0)^(2) =v_(0)^(2) = - 2gh ` [ From equation (1)] ` :. -1/2 v_(0)^(2) = - 2gh` ` :. 2gh = 1/2 v_(0)^(2)` ` :. h = (v_(0)^(2))/(4G) = 49/(4xx9.8)` ` :. h = 1.25 m ` Alternative METHOD : From conservation of mechanical energy `E = K+V` ` :. ` Constant ` = K+V` For at point on ground and at point on height h `K_(1)+V_(1)=K_(2)+V_(2)` `:. K_(1) + 0 = (K_(1))/2 +V_(2)` ` :. V_(2) = K_(1) - (K_(1))/2 ` ` :. V_(2) = (k_(1))/2` ` :."mgh " = 1/2 (1/2 mv_(0)^(2))` ` :. h = (v_(0)^(2))/(4g)` `((7)^(2))/(4xx9.8)` ` :. h = 1.25 m ` |
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