This is a physics problem related to projectile motion. Let's break it down into two components: horizontal and vertical motion.

Known Values:

• Initial horizontal velocity ($v_x$) = 2.5 m/s.
• Time ($t$) = 3.0 s.
• Initial vertical velocity ($v_{y0}$) = 0 m/s (since the diver dives horizontally).
• Gravitational acceleration ($g$) = 9.8 m/s².

1. Height of the cliff (vertical motion):

The vertical motion is governed by the equation of motion for free fall: $y = v_{y0} t + \frac{1}{2} g t^2$ Since $v_{y0} = 0$, this simplifies to: $y = \frac{1}{2} g t^2$ Substitute the values: $y = \frac{1}{2} \times 9.8 \times (3.0)^2$ This will give the height of the cliff.

2. Horizontal distance from the base (horizontal motion):

The horizontal distance is given by: $x = v_x t$ Substitute the values: $x = 2.5 \times 3.0$

Let me calculate the results.The height of the cliff is 44.1 meters, and the diver hit the water 7.5 meters away from the base of the cliff.

Do you need more details or have any questions?

Here are 5 related questions to consider:

1. What would be the time of flight if the height of the cliff doubled?
2. How would the horizontal distance change if the diver's initial speed were 5 m/s?
3. What is the vertical velocity of the diver just before hitting the water?
4. How much time would the diver take to fall from a 60-meter cliff?
5. If air resistance is considered, how would it affect the horizontal distance?

Tip: When solving projectile motion problems, always break the motion into horizontal and vertical components to simplify the calculations.