We analyze the object's motion using the principles of kinematics under constant acceleration, specifically free fall. We can relate the initial velocity, final velocity, acceleration, and displacement without involving time.

[Q1] Let the initial height be $h$. The object starts from rest, so its initial velocity is $v_0 = 0$. The final velocity just before hitting the ground is $v_f = 10$ m/s. The acceleration is due to gravity, $a=g$.

We use the time-independent kinematic equation:

$$v_f^2 = v_0^2 + 2a\Delta y$$

Here, $a = g$ (acceleration due to gravity) and the displacement is the height, $\Delta y = h$. Substituting these into the equation:

$$v_f^2 = v_0^2 + 2gh$$

Since the object starts from rest, $v_0 = 0$:

$$v_f^2 = 2gh$$

We now derive the expression for the initial height $h$:

$$h = \frac{v_f^2}{2g}$$

To find an approximate value, we use the approximation $g \approx 10$ m/s$^2$. Substituting the given values:

$$h = \frac{(10 \text{ m/s})^2}{2(10 \text{ m/s}^2)} = \frac{100 \text{ m}^2/\text{s}^2}{20 \text{ m/s}^2} = 5 \text{ m}$$