This problem involves two-dimensional projectile motion. Since the ball is kicked horizontally, the initial vertical velocity is zero. We'll need to use the horizontal and vertical motion equations separately.

Given:

• Initial horizontal velocity $v_x = 28 \, \text{m/s}$
• Horizontal distance traveled (range) $R = 30 \, \text{m}$
• Acceleration due to gravity $g = 9.8 \, \text{m/s}^2$

We need to find the height of the cliff $h$.

Step 1: Find the time of flight

The horizontal velocity remains constant because there is no acceleration in the horizontal direction (ignoring air resistance). The time of flight $t$ can be calculated using the horizontal motion equation:

$R = v_x \cdot t$

Solving for $t$:

$t = \frac{R}{v_x} = \frac{30}{28} \approx 1.071 \, \text{seconds}$

Step 2: Find the height of the cliff

Now that we know the time of flight, we can calculate the vertical displacement (height of the cliff). The vertical motion is governed by the equation:

$h = \frac{1}{2} g t^2$

Substituting $g = 9.8 \, \text{m/s}^2$ and $t = 1.071 \, \text{seconds}$:

$h = \frac{1}{2} \cdot 9.8 \cdot (1.071)^2 \approx 5.7 \, \text{m}$

Final Answer:

The height of the cliff is approximately 5.7 meters.

Do you want more details or have any questions?

Here are 5 related questions to explore:

1. How does the launch angle affect the time of flight?
2. What would happen if air resistance was considered?
3. How does the initial velocity influence the range of a projectile?
4. Can you calculate the velocity of the ball just before it hits the ground?
5. How does changing the height of the cliff affect the range?

Tip: Always separate horizontal and vertical motions in projectile problems since gravity only affects the vertical motion.