A projectile is launched from ground level at an angle of 35° above the horizontal. It passes through a point that is 15 meters high and 30 meters away horizontally from the launch point. Using only the kinematic equations of projectile motion, find the initial velocity of the projectile.

The horizontal displacement $x$ is given by:

$$x = (v_0 \cos \theta) t$$

The vertical displacement $y$ is given by:

$$y = (v_0 \sin \theta) t - \frac{1}{2} g t^2$$

We are given $x = 30$ m, $y = 15$ m, $\theta = 35^\circ$, and $g = 9.8$ m/s$^2$. We need to find $v_0$.

From the horizontal motion equation:

$$30 = (v_0 \cos 35^\circ) t$$ $$t = \frac{30}{v_0 \cos 35^\circ}$$

$$15 = 30 \frac{\sin 35^\circ}{\cos 35^\circ} - 4.9 \frac{30^2}{(v_0 \cos 35^\circ)^2}$$ $$15 = 30 \tan 35^\circ - \frac{4.9 \times 900}{v_0^2 \cos^2 35^\circ}$$ $$15 = 30 \tan 35^\circ - \frac{4410}{v_0^2 \cos^2 35^\circ}$$

Rearrange to solve for $v_0^2$:

$$\frac{4410}{v_0^2 \cos^2 35^\circ} = 30 \tan 35^\circ - 15$$ $$v_0^2 = \frac{4410}{(30 \tan 35^\circ - 15) \cos^2 35^\circ}$$

Now, calculate the numerical values:
$\tan 35^\circ \approx 0.7002$
$\cos 35^\circ \approx 0.81915$
$\cos^2 35^\circ \approx (0.81915)^2 \approx 0.6709$
Substitute these values:

$$v_0^2 = \frac{4410}{(30 \times 0.7002 - 15) \times 0.6709}$$ $$v_0^2 = \frac{4410}{(21.006 - 15) \times 0.6709}$$ $$v_0^2 = \frac{4410}{(6.006) \times 0.6709}$$ $$v_0^2 = \frac{4410}{4.0306}$$ $$v_0^2 \approx 1094.1$$ $$v_0 = \sqrt{1094.1}$$ $$v_0 \approx 33.077 \text{ m/s}$$

The initial velocity of the projectile is approximately $\text{33.1 m/s}$.