A stunt man jumps horizontally from a building to the roof of a garage that is 2 meters lower. How fast does he need to be to land on the roof of the said garage that is 3 meters away from the building?

To solve this problem, we can use the principles of kinematics and the equation of motion.

Let's assume the initial velocity of the stunt man is v_0 and the time taken to reach the roof of the garage is t. Since the stunt man is jumping horizontally, the vertical component of his velocity is zero.

The horizontal distance traveled (d) is given as 3 meters and the vertical distance (h) is 2 meters.

Using the equation of motion for vertical motion:
h = \frac{1}{2}gt^2
Since the initial vertical velocity is zero, the term involving v_0 disappears.

Simplifying the equation:
2 = \frac{1}{2} \cdot 9.8 \cdot t^2
2 = 4.9 \cdot t^2
t^2 = \frac{2}{4.9}
t^2 \approx 0.4082

Taking the square root of both sides:
t \approx 0.6397 seconds

Now, using the equation of motion for horizontal motion:
d = v_0 \cdot t
Substituting the given values:
3 = v_0 \cdot 0.6397
v_0 = \frac{3}{0.6397}
v_0 \approx 4.69 m/s

Therefore, the stunt man needs to be traveling at approximately 4.69 m/s horizontally to land on the roof of the garage.

\textbf{Answer:} The stunt man needs to be traveling at approximately 4.69 m/s horizontally.