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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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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Answer:
1:7
Explanation:
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