Question

A particle is moving up with balloon with constant acceleration (g/8) which starts from rest from ground and at a height H, a particle is dropped from balloon, After this event, find time for which the particle will be in air (acceleration due to gravity = $gm/s^2$)

• A. $\sqrt{\frac{H}{g}}$
• B. $2\sqrt{\frac{H}{g}}$
• C. $3\sqrt{\frac{H}{g}}$
• D. $4\sqrt{\frac{H}{g}}$

Answer 1

Answer: (B)

$2\sqrt{\frac{H}{g}}$

Answer 2

1. Velocity of balloon at height $H$: $v_p = \sqrt{2 \times \frac{g}{8} \times H} = \sqrt{\frac{gH}{4}} = \frac{\sqrt{gH}}{2}$. This is the initial upward velocity of the dropped particle.
2. Using $y = y_0 + u_0t + \frac{1}{2}at^2$, with $y=0$, $y_0=H$, $u_0=\frac{\sqrt{gH}}{2}$, $a=-g$: $0 = H + \frac{\sqrt{gH}}{2}T - \frac{1}{2}gT^2$
3. Quadratic equation: $\frac{1}{2}gT^2 - \frac{\sqrt{gH}}{2}T - H = 0 \implies gT^2 - \sqrt{gH}T - 2H = 0$.
4. Solving for $T$: $T = \frac{\sqrt{gH} \pm \sqrt{(-\sqrt{gH})^2 - 4(g)(-2H)}}{2g} = \frac{\sqrt{gH} \pm \sqrt{9gH}}{2g} = \frac{4\sqrt{gH}}{2g} = 2\sqrt{\frac{H}{g}}$.