[Q1] Find the height of the balloon above the ground at the moment the object was released.

We analyze the motion of the object after it is released. The object is subject to constant gravitational acceleration. We define a coordinate system with the origin at the ground level and the positive y-direction pointing upwards.

The object's initial velocity, $v_0$, is the same as the balloon's velocity at the moment of release, so $v_0 = 8$ m/s. The acceleration is due to gravity, $a = -g$, where $g = 9.8$ m/s². The object starts at an unknown height $y_0 = h$ and reaches the ground, $y_f = 0$, in time $t = 3$ s.

The kinematic equation for vertical position is:

$$y_f = y_0 + v_0 t + \frac{1}{2} a t^2$$

Substituting the known values into the equation:

$$0 = h + v_0 t + \frac{1}{2} (-g) t^2$$

Now, we derive the expression for the initial height, $h$:

$$h = \frac{1}{2} g t^2 - v_0 t$$

Finally, we substitute the numerical values to find the height:

$$h = \frac{1}{2} (9.8 \, \text{m/s}^2)(3 \, \text{s})^2 - (8 \, \text{m/s})(3 \, \text{s})$$ $$h = \frac{1}{2} (9.8 \, \text{m/s}^2)(9 \, \text{s}^2) - 24 \, \text{m}$$ $$h = 44.1 \, \text{m} - 24 \, \text{m}$$ $$h = 20.1 \, \text{m}$$