To find the distance between the two projectiles in terms of the height \( h \) of the building, we will follow a similar process, but we will use the height \( h \) to derive the necessary time of flight and then relate it to the distance.

### Given:
1. Height of the building \( h \).
2. Initial horizontal velocities \( u_1 \) and \( u_2 \) of the two projectiles.

### Analysis:

1. **Projectile 1** and **Projectile 2** are thrown from the same height \( h \) but in opposite horizontal directions with velocities \( u_1 \) and \( u_2 \) respectively.
2. The velocities of the projectiles will be perpendicular at some time \( t \).

### Vertical Motion Analysis:

Since both projectiles are dropped from the same height, their vertical motion is governed by:
\[ y(t) = h - \frac{1}{2} g t^2 \]

The time \( t \) at which the velocities are perpendicular can be found by setting up the velocity perpendicularity condition as before:
\[ g^2 t^2 = u_1 u_2 \]
\[ t^2 = \frac{u_1 u_2}{g^2} \]
\[ t = \sqrt{\frac{u_1 u_2}{g^2}} = \frac{\sqrt{u_1 u_2}}{g} \]

### Height Relation:

The height \( h \) of the building can be used to find the time of flight to the point where their velocities are perpendicular.

From the vertical motion equation:
\[ h = \frac{1}{2} g t^2 \]

Substituting \( t = \frac{\sqrt{u_1 u_2}}{g} \):
\[ h = \frac{1}{2} g \left( \frac{\sqrt{u_1 u_2}}{g} \right)^2 \]
\[ h = \frac{1}{2} g \cdot \frac{u_1 u_2}{g^2} \]
\[ h = \frac{1}{2} \cdot \frac{u_1 u_2}{g} \]
\[ 2h = \frac{u_1 u_2}{g} \]
\[ \frac{u_1 u_2}{g} = 2h \]

### Horizontal Distance:

Now, the horizontal distance between the projectiles at the time when their velocities are perpendicular is given by:
\[ d_x = (u_1 + u_2) t \]

Using \( t = \frac{\sqrt{u_1 u_2}}{g} \):
\[ d_x = (u_1 + u_2) \cdot \frac{\sqrt{u_1 u_2}}{g} \]

Substitute \( \frac{u_1 u_2}{g} = 2h \):
\[ \sqrt{u_1 u_2} = \sqrt{2hg} \]

Thus:
\[ d_x = (u_1 + u_2) \cdot \frac{\sqrt{2hg}}{g} \]
\[ d_x = (u_1 + u_2) \cdot \sqrt{\frac{2h}{g}} \]

Therefore, the distance between the two projectiles in terms of the height \( h \) of the building when their velocities are perpendicular is:
\[ \boxed{d = (u_1 + u_2) \cdot \sqrt{\frac{2h}{g}}} \]

If you need more details or have any questions, please let me know.

Here are 5 numbered questions for you to ask:
1. How do we derive the time of flight for a projectile given a specific height?
2. Can you explain how the perpendicularity condition affects the time calculation?
3. How can we relate horizontal and vertical motion in projectile problems?
4. Why is the height of the building crucial in this problem?
5. How does the initial velocity impact the distance between projectiles?

Tip: Always check the consistency of units when working through physics problems to ensure your final answer has the correct dimensions.

Learn how to calculate the distance between two projectiles using the height h of a building. This problem involves projectile motion, kinematics, and vertical and horizontal motion equations. Suitable for advanced high school students.

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