Let the launch point be the origin $(0,0)$. The acceleration due to gravity is $g = 9.8$ m/s$^2$.

[Q2] First, find the initial vertical velocity, $v_{0y}$. At the maximum height $\Delta y_{max}$, the vertical velocity is zero. Using the kinematic equation $v_y^2 = v_{0y}^2 + 2a_y \Delta y$:

$$0^2 = v_{0y}^2 - 2g \Delta y_{max}$$ $$v_{0y} = \sqrt{2g \Delta y_{max}}$$ $$v_{0y} = \sqrt{2(9.8 \text{ m/s}^2)(140 \text{ m})} = 52.383 \text{ m/s}$$

[Q1] Next, find the initial horizontal velocity, $v_{0x}$, using the total initial kinetic energy $K_0$.

$$K_0 = \frac{1}{2} m v_0^2 = \frac{1}{2} m (v_{0x}^2 + v_{0y}^2)$$

Solving for $v_{0x}$:

$$v_{0x} = \sqrt{\frac{2K_0}{m} - v_{0y}^2}$$ $$v_{0x} = \sqrt{\frac{2(1550 \text{ J})}{0.55 \text{ kg}} - (52.383 \text{ m/s})^2} = 53.78 \text{ m/s}$$

[Q3] Find the vertical displacement $\Delta y$ when the vertical velocity component is $v_y = 65$ m/s. Since the initial upward velocity is only 52.4 m/s, this velocity must be directed downward, i.e., $v_y = -65$ m/s. Using the same kinematic equation:

$$v_y^2 = v_{0y}^2 - 2g \Delta y$$

Solving for $\Delta y$:

$$\Delta y = \frac{v_{0y}^2 - v_y^2}{2g}$$ $$\Delta y = \frac{(52.383 \text{ m/s})^2 - (65 \text{ m/s})^2}{2(9.8 \text{ m/s}^2)} = -75.56 \text{ m}$$