The speed of a projectile is $v = \sqrt{v_x^2 + v_y^2}$. The horizontal velocity $v_x$ is constant, while the vertical velocity $v_y$ changes due to gravity. The speed is minimum when $v_y = 0$, which occurs at the peak of the trajectory. Therefore, the minimum speed is the horizontal velocity:

$$v_x = v_{min}$$

The time to reach this peak is given as $t_h$. For a symmetric trajectory starting and ending at ground level, the total time of flight is $T = 2t_h$.

The initial speed is $v_0$, which is composed of horizontal and vertical components: $v_0^2 = v_x^2 + v_{y0}^2$. We can find the initial vertical velocity $v_{y0}$:

$$v_{y0} = \sqrt{v_0^2 - v_x^2} = \sqrt{v_0^2 - v_{min}^2}$$

[Q1] The horizontal range $R$ is the constant horizontal velocity multiplied by the total flight time:

$$R = v_x T = v_{min} (2t_h)$$

[Q2] The maximum height $H$ is the vertical distance traveled in time $t_h$. Using the kinematic equation $H = \frac{v_{y0} + v_y}{2}t$, where $v_y = 0$ at the peak:

$$H = \frac{v_{y0} + 0}{2} t_h = \frac{1}{2}v_{y0}t_h$$

Substituting the expression for $v_{y0}$:

$$H = \frac{1}{2}t_h\sqrt{v_0^2 - v_{min}^2}$$