We can solve this problem using the principles of projectile motion. First, let's break down the given information:

• Initial velocity ($v_0$) = 25 m/s
• Angle of projection ($\theta$) = 30°
• Acceleration due to gravity ($g$) = 9.8 m/s²
• Initial height (since it's shot from a cliff, we'll assume this is non-zero unless specified; if we assume it's zero, it will be a standard projectile problem)

To find each part:

a. Time of Flight

The total time the projectile is in the air can be found using:

$t = \frac{2 v_0 \sin(\theta)}{g}$

Where $v_0 \sin(\theta)$ is the vertical component of the initial velocity.

b. How far does it go? (Range)

The horizontal distance or range is given by:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

c. Final Velocity

The final velocity of the ball consists of both horizontal and vertical components. Using the equations for each:

• Horizontal velocity: $v_{x} = v_0 \cos(\theta)$
• Vertical velocity: $v_y = v_0 \sin(\theta) - g t$

The magnitude of the final velocity ($v_f$) is found by combining these components:

$v_f = \sqrt{v_x^2 + v_y^2}$

Let me calculate the results using these equations.Here are the results for the ball's motion:

a. Time of flight: 2.55 seconds

b. Range: The ball travels 55.23 meters horizontally.

c. Final velocity: The final velocity of the ball is 25.0 m/s (the same as the initial velocity, because the vertical and horizontal components combine to give the same magnitude).

Would you like further details on any of these calculations?

Here are 5 related questions for further exploration:

1. What happens to the range if the angle is increased to 45°?
2. How does the time of flight change with different initial velocities?
3. What if the cliff's height was 20 meters—how would the equations change?
4. What is the maximum height the ball reaches?
5. How would air resistance affect the ball's motion?

Tip: In projectile motion, the horizontal velocity remains constant (ignoring air resistance), while the vertical velocity changes due to gravity.