The motion is analyzed by separating it into horizontal (x) and vertical (y) components. Initial conditions: $v_{0x} = v_0$ and $v_{0y} = 0$. Acceleration: $a_x = 0$ and $a_y = g$ (taking downward as the positive direction).

The final velocity has magnitude $v_f$ at an angle $\theta$ with the horizontal. The components of the final velocity are:

$$v_{fx} = v_f \cos(\theta)$$ $$v_{fy} = v_f \sin(\theta)$$

Since horizontal velocity is constant, the initial speed is $v_0 = v_{fx} = v_f \cos(\theta)$.

[Q1] Find the initial height from which the object was launched. We use the time-independent kinematic equation for the vertical motion:

$$v_{fy}^2 = v_{0y}^2 + 2gh$$

Substituting $v_{0y} = 0$ and the expression for $v_{fy}$:

$$(v_f \sin(\theta))^2 = 0^2 + 2gh$$

Solving for the initial height $h$:

$$h = \frac{v_f^2 \sin^2(\theta)}{2g}$$

[Q2] Find the horizontal range of the object. The horizontal range is $R = v_{0x}t$. We first need to find the time of flight, $t$. From the vertical motion kinematics:

$$v_{fy} = v_{0y} + gt$$

Substituting known values and solving for $t$:

$$v_f \sin(\theta) = 0 + gt$$ $$t = \frac{v_f \sin(\theta)}{g}$$

Now, substitute $v_{0x}$ and $t$ into the range equation:

$$R = (v_f \cos(\theta)) \left(\frac{v_f \sin(\theta)}{g}\right)$$ $$R = \frac{v_f^2 \sin(\theta)\cos(\theta)}{g}$$

Using the trigonometric identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, this can also be written as:

$$R = \frac{v_f^2 \sin(2\theta)}{2g}$$