A man in a balloon rising vertically with an acceleration of \(4.9 \, m / sec^{2}\) releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is \(\left(g = 9.8\, m / \sec^{2}\right)\)
Text Solution
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Height travelled by ball (with balloon) in 2 sec
\(h_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 4.9 \times 2^2 = 9.8\,m\)
Velocity of the balloon after 2 sec
\(v = at = 4.9 \times 2 = 9.8 \, m/s\)
Now if the ball is released from the balloon then it acquire same velocity in upward direction.
Let it move up to maximum height h_2
v^{2} = u^{2} = 2gh_{2} ⇒ ⇒ \(0 = (9.8)^2 - 2 \times (9.8) \times h_2\) 
h_2 =4.9 m
Greatest height above the ground reached by the ball \(= h_{1} + h_{2} = 9.8 + 4.9 = 14.7\,m\)
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