To solve this problem, we can break the motion into horizontal and vertical components, using kinematic equations. We'll work through the following steps:

Step 1: Break Down the Motion

• Horizontal Component (x-axis): $v_{x} = v_0 \cdot \cos(\theta)$ where $v_0$ is the initial velocity and $\theta = 25^\circ$.

• Vertical Component (y-axis): $v_{y} = v_0 \cdot \sin(\theta)$

Step 2: Horizontal Motion Equation

The horizontal motion is given by: $x = v_x \cdot t$ where $x = 25 \, \text{m}$ (the horizontal distance) and $t$ is the time of flight.

Step 3: Vertical Motion Equation

For the vertical motion, we use the kinematic equation: $y = v_y \cdot t - \frac{1}{2} g t^2$ where $y = -20 \, \text{m}$ (since the ball lands 20 meters below the starting point) and $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.

Step 4: Solving the Equations

1. Express the time of flight $t$ from the horizontal motion equation: $t = \frac{x}{v_x} = \frac{25}{v_0 \cdot \cos(25^\circ)}$
2. Substitute this expression for $t$ into the vertical motion equation: $-20 = v_0 \cdot \sin(25^\circ) \cdot \frac{25}{v_0 \cdot \cos(25^\circ)} - \frac{1}{2} \cdot 9.8 \cdot \left(\frac{25}{v_0 \cdot \cos(25^\circ)}\right)^2$
3. Solve for $v_0$.

Let me do the math to find the initial speed $v_0$.The initial speed of the ball is approximately $14.89 \, \text{m/s}$.

This result is derived from solving the kinematic equations for the projectile's motion, taking into account both horizontal and vertical components.

Would you like further clarification on any part of the process? Here are some related questions:

1. How does the angle of projection affect the range of the projectile?
2. What would happen if the angle of projection were increased to 45°?
3. How do we determine the time of flight of the projectile?
4. What other factors could influence the projectile's motion in a real-world scenario?
5. How would air resistance alter the calculations?

Tip: The initial speed and angle of projection determine the entire trajectory of a projectile, influencing both the range and the maximum height.