Question

A man in a balloon rising vertically with an acceleration of $4.9m/\sec^{2}{}$ 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 $(g = 9.8m/\sec^{2})$

• A. 14.7 m
• B. 19.6 m
• C. 9.8 m
• D. 24.5 m

Answer 1

Answer: (A)

14.7 m

Answer 2

Height travelled by ball (with balloon) in 2 sec

$h_{1} = \frac{1}{2}a\mspace{6mu} t^{2} = \frac{1}{2} \times 4.9 \times 2^{2} = 9.8\mspace{6mu} m$

Velocity of the balloon after 2 sec

$v = a\mspace{6mu} t = 4.9 \times 2 = 9.8\mspace{6mu} m/s$

Now if the ball is released from the balloon then it acquire same velocity in upward direction.

Let it move up to maximum height $h_{2}$

$v^{2} = u^{2} - 2gh_{2}$ ⇒ $0 = (9.8)^{2} - 2 \times (9.8) \times h_{2}$ $\therefore$ $h_{2}$=4.9m

Greatest height above the ground reached by the ball$= h_{1} + h_{2} = 9.8 + 4.9 = 14.7\mspace{6mu} m$