Problem Breakdown

We are given:

• Initial horizontal velocity $v_x = 45 \, \text{m/s}$
• Height from which the cannonball is launched $h = 20 \, \text{m}$
• The vertical motion is subject to gravity with $g = 9.8 \, \text{m/s}^2$
• Horizontal motion is constant because there’s no air resistance.

Let's solve the problem step-by-step:

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Part A: Time to reach the ground

We know the vertical distance the cannonball falls is 20 meters. We can use the following kinematic equation for vertical motion:

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

Solving for $t$, the time to reach the ground:

$20 = \frac{1}{2} (9.8) t^2$

$t^2 = \frac{20}{4.9}$

$t^2 = 4.08$

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

So, the cannonball will take 2.02 seconds to reach the ground.

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Part B: Range of the cannonball

The horizontal distance (range) is calculated using the formula:

$\text{Range} = v_x \times t$

Substitute the known values:

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

So, the range of the cannonball is 90.9 meters.

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Part C: Magnitude of velocity when it strikes the ground

The velocity when the cannonball strikes the ground is the resultant of its horizontal and vertical velocity components.

• The horizontal velocity remains constant at $v_x = 45 \, \text{m/s}$.
• The vertical velocity $v_y$ at the moment it strikes the ground is given by:

$v_y = g \times t = 9.8 \, \text{m/s}^2 \times 2.02 \, \text{seconds} \approx 19.8 \, \text{m/s}$

Now, the total velocity $v$ is the resultant of $v_x$ and $v_y$:

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

$v = \sqrt{45^2 + 19.8^2}$

$v = \sqrt{2025 + 392.04} \approx \sqrt{2417.04} \approx 49.16 \, \text{m/s}$

So, the magnitude of the velocity when it strikes the ground is approximately 49.16 m/s.

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Part D: Direction of the velocity when it strikes the ground

The direction (or angle $\theta$) at which the cannonball strikes the ground can be found using the following relation:

$\tan(\theta) = \frac{v_y}{v_x}$

$\tan(\theta) = \frac{19.8}{45}$

$\theta = \tan^{-1}\left(\frac{19.8}{45}\right) \approx \tan^{-1}(0.44)$

$\theta \approx 23.74^\circ$

Thus, the cannonball strikes the ground at an angle of 23.74° below the horizontal.

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Summary of Answers:

• (a) Time to reach the ground: 2.02 seconds
• (b) Range of the cannonball: 90.9 meters
• (c) Magnitude of the velocity: 49.16 m/s
• (d) Direction of the velocity: 23.74° below the horizontal

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Do you want further details or have any questions?

Here are 5 related questions:

1. What if the height of the building was doubled? How would the time change?
2. How does air resistance affect the range and velocity?
3. Can we calculate the maximum range of a cannonball given an initial angle?
4. How do different gravitational accelerations affect the result (e.g., on the Moon)?
5. What happens to the direction of the velocity if the initial horizontal speed is decreased?

Tip: When analyzing projectile motion, always treat horizontal and vertical motions separately.