Q. A balloon starts rising from the ground with an acceleration of $1.25\ m/s^{2}$ after 8s, a stone is released from the balloon. The stone will $(g - 10\ m/s^{2})$
A) Reach the ground in 4 second
B) Begin to move down after being released
C) Have a displacement of $50\ m$
D) Cover a distance of $40\ m$ in reaching the ground
Explanation
(a) When the stone is released from the balloon. Its height
$h = \frac{1}{2}at^{2} = \frac{1}{2} \times 1.25 \times (8)^{2} = 40\ m$ and velocity
$v = at = 1.25 \times 8 = 10\ m/s$
Time taken by the stone to reach the ground
$t = \frac{v}{g}\left\lbrack 1 + \sqrt{1 + \frac{2gh}{v^{2}}} \right\rbrack = \frac{10}{10}\left\lbrack 1 + \sqrt{1 + \frac{2 \times 10 \times 40}{(10)^{2}}} \right\rbrack = 4sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.