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A man in a balloon rising vertically with an acceleration of 5ms−2 releases a ball 2 second after the balloon is let go from the ground. The greatest height above the ground reached by the ball is : |
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Answer» Height traveled by ball (with balloon) in 2 sec h1=1/2at2=1/2×5×22= 10 m Velocity of the balloon after 2 sec v = at = 5×2 = 10 m/s Now if the ball is released from the balloon then it acquire same velocity in upward direction. Let it move up to maximum height h2 v2=u2−2gh2 => 0=(10)2−2×(10)×h2 ∴ h2 = 5m Greatest height above the ground reached by the ball = h1+h2 = 10+5 = 15 m |
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