height-left-greatest-frac-ground-ballon

A man in a ballon rising vertically with an acceleration of

4.9 \frac{m}{s e c^{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

(g = \frac{9.8 m}{s e c^{2}})

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 = a t = 4.9 \times 2 = 9.8 \frac{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} - 2 g h_{2} \Rightarrow 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