two projectiles are thrown from a tower in opposite direction with u1 and u2 find the distance btwn balls when the velocity are perpendicular(find distance in terms of height of building)

To find the distance between the two projectiles when their velocities are perpendicular, we need to analyze the motion of both projectiles and set up the equations accordingly.

### Given:
1. Two projectiles are thrown from a tower in opposite directions with initial velocities \( u_1 \) and \( u_2 \).
2. We need to find the distance between the two projectiles when their velocities are perpendicular.
3. Let \( h \) be the height of the building.

### Analysis:
1. **Projectile 1** is thrown with initial velocity \( u_1 \) at an angle \( \theta_1 \) to the horizontal.
2. **Projectile 2** is thrown with initial velocity \( u_2 \) at an angle \( \theta_2 \) to the horizontal, in the opposite direction.

For simplicity, we will assume both projectiles are thrown horizontally (angles \( \theta_1 = 0^\circ \) and \( \theta_2 = 180^\circ \)). This simplifies the initial conditions to horizontal velocities \( u_1 \) and \( u_2 \).

### Equations of Motion:
The motion of each projectile can be described by the following equations:

For projectile 1:
\[ x_1 = u_1 t \]
\[ y_1 = h - \frac{1}{2} g t^2 \]

For projectile 2:
\[ x_2 = -u_2 t \]
\[ y_2 = h - \frac{1}{2} g t^2 \]

Here, \( x_1 \) and \( x_2 \) are the horizontal displacements, \( y_1 \) and \( y_2 \) are the vertical displacements from the top of the tower, and \( g \) is the acceleration due to gravity.

### Velocity Components:
The velocities of the projectiles at any time \( t \) are given by:

For projectile 1:
\[ v_{x1} = u_1 \]
\[ v_{y1} = -g t \]

For projectile 2:
\[ v_{x2} = -u_2 \]
\[ v_{y2} = -g t \]

### Condition for Perpendicular Velocities:
The velocities are perpendicular when the dot product of the velocity vectors is zero:
\[ \vec{v_1} \cdot \vec{v_2} = 0 \]

\[ (u_1, -g t) \cdot (-u_2, -g t) = 0 \]

\[ -u_1 u_2 + g^2 t^2 = 0 \]

\[ g^2 t^2 = u_1 u_2 \]

\[ t = \sqrt{\frac{u_1 u_2}{g^2}} \]

\[ t = \sqrt{\frac{u_1 u_2}{g^2}} = \sqrt{\frac{u_1 u_2}{g}} \]

### Distance Between the Projectiles:
At this time \( t \), the distance between the projectiles is given by the Pythagorean theorem considering their horizontal and vertical displacements:

Horizontal distance:
\[ d_x = x_1 - x_2 = u_1 t + u_2 t = t (u_1 + u_2) \]

Vertical distance:
\[ d_y = 0 \] (since they are at the same vertical height \( y \))

The distance \( D \) between the projectiles is:
\[ D = \sqrt{d_x^2 + d_y^2} = d_x \]

Substituting \( t \):
\[ D = t (u_1 + u_2) = \sqrt{\frac{u_1 u_2}{g}} (u_1 + u_2) \]

Thus, the distance between the projectiles when their velocities are perpendicular is:
\[ D = \sqrt{\frac{u_1 u_2}{g}} (u_1 + u_2) \]

### Conclusion:
The distance between the two projectiles when their velocities are perpendicular is:
\[ D = \sqrt{\frac{u_1 u_2}{g}} (u_1 + u_2) \]

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 you derive the equation of motion for a projectile?
2. Can you explain how to find the horizontal and vertical components of a projectile's velocity?
3. How does the condition for perpendicular velocities relate to the dot product of vectors?
4. What assumptions are made in this projectile motion problem?
5. How does the acceleration due to gravity influence the motion of projectiles?

Tip: When solving problems involving projectile motion, always start by breaking the motion into horizontal and vertical components and applying the appropriate kinematic equations to each direction.

Learn how to find the distance between two projectiles thrown in opposite directions from a tower when their velocities are perpendicular. This problem involves analyzing projectile motion, kinematic equations, and applying the Pythagorean theorem. Suitable for high school students studying physics and projectile motion.

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