To solve the problem, we'll use the principles of projectile motion. The problem can be broken down into two parts: vertical motion (due to gravity) and horizontal motion (which is constant since there is no air resistance).

Given Data:

• Height of the building (h): 20 meters
• Initial horizontal velocity (v₀ₓ): 45 m/s
• Acceleration due to gravity (g): 9.8 m/s²

(a) Time to reach the ground

To find the time it takes for the cannonball to reach the ground, we'll analyze the vertical motion. The vertical motion is influenced by gravity and can be described by the following kinematic equation:

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

Solving for time $t$:

$t = \sqrt{\frac{2h}{g}}$

Substituting the given values:

$t = \sqrt{\frac{2 \times 20 \, \text{m}}{9.8 \, \text{m/s}^2}}$

Let's calculate this.

$t = \sqrt{\frac{40}{9.8}} \approx \sqrt{4.08} \approx 2.02 \, \text{seconds}$

So, the time it takes for the cannonball to reach the ground is approximately 2.02 seconds.

(b) Range of the cannonball

The range of the cannonball is determined by the horizontal distance it travels during the time it is in the air. Since there is no horizontal acceleration, the horizontal distance (range) can be calculated using:

$\text{Range} = v₀ₓ \times t$

Substituting the values:

$\text{Range} = 45 \, \text{m/s} \times 2.02 \, \text{seconds} \approx 90.9 \, \text{meters}$

Thus, the range of the cannonball as it reaches the ground is approximately 90.9 meters.

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Do you want more details or have any questions about this solution?

Here are 5 related questions to expand your understanding:

1. What would happen to the range if the initial speed was doubled?
2. How would the time to reach the ground change if the building were taller?
3. What is the significance of neglecting air resistance in projectile motion problems?
4. How would the range be affected if the cannonball was launched at an angle instead of horizontally?
5. If the initial velocity had a vertical component, how would the time to hit the ground be affected?

Tip: In projectile motion problems, always separate the vertical and horizontal components and solve them individually using the appropriate kinematic equations.