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Let the reference level as given. Here the potential energy be zero.
From energy conservation
Potential Energy at B = K.E at A
Mgh = 1/2 MV2 ......................................... Where V is the velocity at A
Mgh - 5% Mgh = 1/2 MV2 ....................... Given
gh - 5% gh = 1/2 V2 .................................... (As 5% of energy is lost against air resistance)
gh - 5/100gh = 1/2 V2
95/100 gh = 1/2 V2
V2 = 190/100 gh
Thus bob comes down with speed of 5.33 m/s
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