Step-by-Step Solution

1. Calculate the initial vertical and horizontal velocities:

The initial velocity ($v_0 = 12.4 \, \text{m/s}$) is split into horizontal ($v_{0x}$) and vertical ($v_{0y}$) components using trigonometric functions:

$v_{0x} = v_0 \cos\theta, \quad v_{0y} = v_0 \sin\theta$

$v_{0x} = 12.4 \cos(60^\circ) = 12.4 \cdot 0.5 = 6.2 \, \text{m/s}$

$v_{0y} = 12.4 \sin(60^\circ) = 12.4 \cdot 0.866 = 10.74 \, \text{m/s}$

2. Time to land:

The total time of flight can be determined using the vertical motion. The formula for the time to reach the maximum height is:

$t_\text{up} = \frac{v_{0y}}{g}$

$t_\text{up} = \frac{10.74}{9.8} = 1.096 \, \text{s}$

The total time of flight is twice the time to reach the maximum height because the motion is symmetric:

$t_\text{total} = 2 \cdot t_\text{up} = 2 \cdot 1.096 = 2.192 \, \text{s}$

3. Maximum height:

The maximum height is reached when the vertical velocity becomes zero. The formula for the maximum height is:

$h_\text{max} = \frac{v_{0y}^2}{2g}$

$h_\text{max} = \frac{10.74^2}{2 \cdot 9.8} = \frac{115.32}{19.6} = 5.89 \, \text{m}$

4. Horizontal distance (range):

The horizontal distance is given by:

$R = v_{0x} \cdot t_\text{total}$

$R = 6.2 \cdot 2.192 = 13.59 \, \text{m}$

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Final Results

1. Initial Velocities:

   • Horizontal velocity ($v_{0x}$) = $6.2 \, \text{m/s}$
   • Vertical velocity ($v_{0y}$) = $10.74 \, \text{m/s}$

2. Time to land: $2.19 \, \text{seconds}$

3. Maximum height: $5.89 \, \text{meters}$

4. Horizontal range: $13.59 \, \text{meters}$

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Would you like to see detailed plots or explanations for any of these steps?

Expanding Questions:

1. How does the angle of launch affect the range and maximum height?
2. What would happen if the cannonball were fired at a different speed?
3. How would air resistance alter the results of this problem?
4. What other factors could affect projectile motion in real-life scenarios?
5. How can you derive these equations from the basic kinematic equations?

Tip:

For any projectile problem, always start by resolving the velocity into its horizontal and vertical components. This simplifies calculations and avoids confusion!