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A ball is thrown downward from a height of 30 m with a velocityof 10 m s1. Determine the velocity with which the ball strikesthe ground by using law of conservation of energy |
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Answer» P.E Initial = mgh = m*(10*30)= 300mK.E initial = 1/2*m*v² = 50m now finally , P.E = 0 , K.E = m*v²/2 so, by Conservation of energy mv²/2 = 300m+50m=> v² = 2*350 = 700=> v = 10√7 m/s |
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