Let the initial velocity be $\vec{v}_0$ with magnitude $v_0$ and angle $\theta$ above the horizontal. The initial velocity components are $v_{0x} = v_0 \cos\theta$ and $v_{0y} = v_0 \sin\theta$. The acceleration due to gravity is $g$, acting downwards.

The motion is decomposed into horizontal (constant velocity) and vertical (constant acceleration) components. Horizontal motion: $v_x(t) = v_{0x}$ Vertical motion: $v_y(t) = v_{0y} - gt$

At the maximum height, the vertical component of velocity is zero, $v_y = 0$. This occurs at time $t$. The speed at this point, $v_h$, is purely horizontal.

$$v_h = v_x(t) = v_{0x}$$

[Q1] Find the magnitude of the object's initial velocity. From the vertical motion equation at time $t$:

$$0 = v_{0y} - gt \implies v_{0y} = gt$$

The magnitude of the initial velocity, $v_0$, is found by combining its horizontal and vertical components using the Pythagorean theorem:

$$v_0 = \sqrt{v_{0x}^2 + v_{0y}^2}$$

Substituting the expressions for $v_{0x}$ and $v_{0y}$:

$$v_0 = \sqrt{v_h^2 + (gt)^2}$$

[Q2] Find the maximum height the object reaches. The maximum height, $H$, is the vertical displacement at time $t$. Using the kinematic equation for vertical displacement:

$$H = v_{0y}t - \frac{1}{2}gt^2$$

Substitute $v_{0y} = gt$ from the previous step:

$$H = (gt)t - \frac{1}{2}gt^2 = gt^2 - \frac{1}{2}gt^2$$ $$H = \frac{1}{2}gt^2$$

[Q3] Find the range the object reaches. The range, $R$, is the total horizontal distance traveled. Assuming the object lands at the same vertical level it was launched from, the total time of flight, $T$, is twice the time to reach the maximum height due to the symmetry of the trajectory.

$$T = 2t$$

The range is the product of the constant horizontal velocity and the total time of flight:

$$R = v_{0x} T$$

Substituting $v_{0x} = v_h$ and $T = 2t$:

$$R = v_h (2t) = 2v_h t$$