Understanding the Problem: Time Taken by the Body Dropped from a Balloon

The question asks for the time taken by a body to reach the ground after being dropped from a balloon. The balloon is moving upward with a uniform acceleration. The body is dropped after a certain time, inheriting the balloon's velocity at that instant.

Analyzing Balloon Motion

First, let's analyze the motion of the balloon before the body is dropped. The balloon starts from rest and moves upward with a uniform acceleration $a_b = \frac{g}{8}$. The body is dropped after half a minute, which is $t_1 = 30$ seconds.

At the moment the body is dropped (after 30 seconds), the balloon has a certain upward velocity and is at a certain height. This velocity and height will be the initial conditions for the dropped body.

• Initial velocity of balloon, $u_b = 0$ (assuming it starts from rest).
• Acceleration of balloon, $a_b = \frac{g}{8}$.
• Time duration, $t_1 = 30$ seconds.

The velocity of the balloon at $t_1 = 30$ seconds can be found using the first kinematic equation:

$\qquad v_b = u_b + a_b t_1$

$\qquad v_b = 0 + \left(\frac{g}{8}\right) \times 30$

$\qquad v_b = \frac{30g}{8} = \frac{15g}{4}$ m/s (upward)

The height of the balloon at $t_1 = 30$ seconds can be found using the second kinematic equation:

$\qquad h = u_b t_1 + \frac{1}{2} a_b t_1^2$

$\qquad h = (0)(30) + \frac{1}{2} \left(\frac{g}{8}\right) (30)^2$

$\qquad h = \frac{1}{16} g \times 900$

$\qquad h = \frac{900g}{16} = \frac{225g}{4}$ meters

So, when the body is dropped, it is at a height of $\frac{225g}{4}$ meters above the ground, and it has an initial upward velocity of $\frac{15g}{4}$ m/s, inherited from the balloon motion.

Initial Conditions for the Dropped Body

When the body is dropped, it is no longer under the influence of the balloon's acceleration. It is now a dropped body moving under the influence of gravity alone.

• Initial velocity of the dropped body, $u_{body} = \frac{15g}{4}$ m/s (upward, positive direction).
• Initial height of the dropped body from the ground, $h = \frac{225g}{4}$ meters.
• Acceleration acting on the dropped body, $a_{body} = -g$ m/s$^2$ (downward, negative direction).

We want to find the time taken by the body to reach the ground. The final position is the ground, so the displacement of the body is downward from its initial height. Therefore, the displacement $s_{body} = -h = -\frac{225g}{4}$ meters.

Applying Kinematic Equation to Find Time Taken by the Body

We use the kinematic equation relating displacement, initial velocity, acceleration, and time:

$\qquad s_{body} = u_{body} t_2 + \frac{1}{2} a_{body} t_2^2$

Substitute the known values:

$\qquad -\frac{225g}{4} = \left(\frac{15g}{4}\right) t_2 + \frac{1}{2} (-g) t_2^2$

We can divide the entire equation by $g$ (assuming $g \neq 0$):

$\qquad -\frac{225}{4} = \frac{15}{4} t_2 - \frac{1}{2} t_2^2$

To eliminate the fractions, multiply the entire equation by 4:

$\qquad -225 = 15 t_2 - 2 t_2^2$

Rearrange the terms to form a standard quadratic equation $At_2^2 + Bt_2 + C = 0$:

$\qquad 2 t_2^2 - 15 t_2 - 225 = 0$

Calculating Time Taken by the Body: Solving the Quadratic Equation

We use the quadratic formula to solve for $t_2$:

$\qquad t_2 = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

Here, $A=2$, $B=-15$, and $C=-225$.

$\qquad t_2 = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(2)(-225)}}{2(2)}$

$\qquad t_2 = \frac{15 \pm \sqrt{225 + 1800}}{4}$

$\qquad t_2 = \frac{15 \pm \sqrt{2025}}{4}$

The square root of 2025 is 45.

$\qquad t_2 = \frac{15 \pm 45}{4}$

This gives two possible values for $t_2$:

$\qquad t_{2,1} = \frac{15 + 45}{4} = \frac{60}{4} = 15$ seconds

$\qquad t_{2,2} = \frac{15 - 45}{4} = \frac{-30}{4} = -7.5$ seconds

Since time cannot be negative, the physically meaningful solution for the time taken by the body to reach the ground is 15 seconds.

This calculation confirms the time taken by the body to reach the ground is 15 seconds after being dropped.

Summary of Dropped Body Motion

The dropped body initially moves upward due to the inherited velocity from the balloon, reaches its highest point, and then falls back down to the ground under gravitational acceleration. The total time for this path is 15 seconds.

The time taken by the body to reach the ground is indeed 15 seconds.