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. Calculate the velocity of the balloon when it reaches height $H$ using $v^2 = u^2 + 2as$. With $u=0$, $a=g/8$, and $s=H$, the velocity is $v_{balloon} = \sqrt{2(g/8)H} = \frac{\sqrt{gH}}{2}$.
2. When the particle is dropped, its initial velocity is $u_0 = v_{balloon} = \frac{\sqrt{gH}}{2}$ (upwards), initial position is $y_0 = H$, and acceleration is $a = -g$.
3. Use the kinematic equation $y = y_0 + u_0t + \frac{1}{2}at^2$ to find the time $t$ when the particle hits the ground ($y=0$).
4. The equation becomes $0 = H + \frac{\sqrt{gH}}{2}t - \frac{1}{2}gt^2$. Solving this quadratic equation for $t$ yields the positive time $t = 2\sqrt{\frac{H}{g}}$.