To find the height of the cliff, we need to determine the vertical distance the cannonball falls over the 2-second time period. Since the cannonball is fired horizontally, there is no initial vertical velocity; only the effect of gravity acts on it in the vertical direction.

The vertical distance $h$ can be calculated using the formula:

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

where:

• $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
• $t = 2 \, \text{s}$ (time it takes to hit the ground).

Step-by-Step Solution

1. Substitute the values into the formula:

   $h = \frac{1}{2} \times 9.8 \times (2)^2$

2. Calculate the time squared:

   $h = \frac{1}{2} \times 9.8 \times 4$

3. Multiply the terms:

   $h = \frac{1}{2} \times 39.2$

4. Solve for $h$:

   $h = 19.6 \, \text{meters}$

Conclusion

The height of the cliff is 19.6 meters.

Would you like further details or have any other questions?

---

1. How would the height change if the cannonball took 3 seconds to hit the ground?
2. What would happen to the range if the initial speed increased to 15 meters per second?
3. How is horizontal velocity calculated in projectile motion problems?
4. How do you determine time of flight for an object dropped from a certain height?
5. How would air resistance affect the vertical motion in this scenario?

Tip: In projectile motion problems, separating horizontal and vertical components often simplifies calculations.