Step 1: Understand the condition for Maximum Elevation Angle
Let the coordinates of the balloon at time $t$ relative to the telescope be $(x, y)$. The angle of elevation $\theta$ is given by: $$ \tan \theta = \frac{y}{x} $$ For $\theta$ to be maximum, $\frac{d\theta}{dt} = 0$. Differentiating $\tan \theta = \frac{y}{x}$ with respect to time: $$ \sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{x \frac{dy}{dt} – y \frac{dx}{dt}}{x^2} $$ Setting $\frac{d\theta}{dt} = 0$, the numerator must be zero: $$ x v_y – y v_x = 0 \implies v_x = v_y \frac{x}{y} $$ $$ v_x = \frac{v_y}{\tan \theta} $$
Note: Geometrically, this means the net velocity vector is aligned with the position vector (line of sight).

Step 2: Substitute values
From the graph, the maximum angle $\theta_{max} = 53^\circ$.
From Question 29, $v_y = 8 \text{ m/min}$.
$$ v_x = \frac{8}{\tan(53^\circ)} $$ Using the standard triangle for $53^\circ$, $\tan(53^\circ) \approx \frac{4}{3}$. $$ v_x = \frac{8}{4/3} $$ $$ v_x = 8 \times \frac{3}{4} $$ $$ v_x = 6 \text{ m/min} $$