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A body is thrown vertically upward with velocity u. The ratio of times at which it is at particular height h is 1:2. What is the maximum height reached by body above point of projection? |
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Answer» Le H is the maximum height. LET h is the height above the ground at which both P.E and K.E are equal ∴ T e at the point p = K.E + P.E ⇒ K.E = mgH - mgh = mg [H - h] ∴ P.E / K.E = mgh/mg[HH] As K.E = P.E ∴ h / H - h = 1, H – h = 2h …(1) When the body is projected upwards with VELOCITY u it reaches a maximum height H ∴ H = u2/ 2g When the body is projected vertically upwards with double the velocity then it reaches a maximum height H1 ∴ H = (2u)2/2g = 4u2/2g = 4H………..(2) ∴ The ratio of P.E and K.E at the same point ‘P’ = P.E / K.E = mgh/mg[H1-h] ∴ P.E / K.E = h/4[2h]-h = 1/7 Or, 1:7 Hope this helps you buddy :) |
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