Question

A man in a balloon rising vertically with an acceleration of 4.9m/s^{​​​​​​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(g=9.8)

Answer 1

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Height travelled by ball (with balloon) in 2 sec
h1=12at2=12×4.9×22=9.8m

Velocity of the balloon after 2 sec

v=at=4.9×2=9.8ms

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=(9.8)2−2×(9.8)×h2

∴h2=4.9m

Greatest height above the ground reached by the ball = h1+h2=9.8+4.9=14.7m