Step 1: Analyze the initial motion
From the problem statement, we know the wind velocity is approximately zero up to a height of $30 \text{ m}$. During this phase, the balloon rises vertically. The graph shows that the motion starts at $t = 3.35 \text{ min}$ with an initial angle of elevation $\theta = 15^\circ$.

Let the ascent velocity be $v_y$. Since the balloon moves vertically from the release point (which is at a horizontal distance $l = 100 \text{ m}$ from the telescope), we can form a right-angled triangle.

Step 2: Calculate vertical height ($h$)
Using the right-angled triangle formed by the ground, the vertical path, and the line of sight:
$$ \tan(15^\circ) = \frac{h}{l} = \frac{h}{100} $$ Using the value $\tan(15^\circ) \approx 0.268$ (calculated as $\tan(45^\circ-30^\circ)$): $$ h = 100 \times 0.268 = 26.8 \text{ m} $$

Step 3: Calculate Ascent Velocity ($v_y$)
The balloon reached this height at $t = 3.35 \text{ min}$. Since ascent velocity is constant: $$ v_y = \frac{h}{t} $$ $$ v_y = \frac{26.8}{3.35} $$ $$ v_y = 8 \text{ m/min} $$