Given the problem, you're firing an object from a cannon at an angle of 25 degrees with an initial horizontal velocity of 10 m/s and starting at ground level (0 m height). To calculate the horizontal displacement (or range), we need to consider the object's horizontal motion and the time it spends in the air.

Assumptions:

• There is no air resistance.
• The gravitational acceleration $g = 9.8 \, \text{m/s}^2$.

Step-by-Step Solution:

1. Initial Velocity Components: The initial velocity has both horizontal and vertical components:

   • Horizontal velocity $v_x = 10 \, \text{m/s}$ (Given)
   • Vertical velocity $v_y = v_0 \sin(\theta) = 10 \sin(25^\circ) \approx 4.23 \, \text{m/s}$

2. Time of Flight: The time of flight can be determined using the vertical motion of the object. Since it starts and ends at the same height (0 m), the formula for the time of flight is:

   $t = \frac{2 v_y}{g} = \frac{2 \times 4.23}{9.8} \approx 0.86 \, \text{seconds}$

3. Horizontal Displacement: The horizontal displacement is calculated using the horizontal velocity and the time of flight:

   $x = v_x \times t = 10 \times 0.86 \approx 8.6 \, \text{m}$

Answer:

The object is displaced approximately 8.6 meters in the x-direction (horizontal distance).

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Would you like any further details or explanations?

Here are five related questions you might find interesting:

1. How would the displacement change if the initial velocity was doubled?
2. How does air resistance affect the horizontal displacement in real-world scenarios?
3. What would happen to the time of flight if the angle were increased to 45 degrees?
4. How do we account for the height of the cannon in similar problems?
5. Can we derive a general formula for the range of a projectile?

Tip: When launching projectiles, the horizontal distance is maximized at a 45-degree angle under ideal conditions (without air resistance).