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Projectile motion involves analyzing the initial velocity of an object, which is delivered in two directions: horizontal and vertical. These directions are aptly referred to as the "components" of the initial velocity.
When a boy throws a ball from a roof, it's launched at a speed that's split between these two components. First, we calculate the vertical component using the formula:

• \( u_y = u \cdot \sin(\theta) \)
• where \( u \) is the initial speed and \( \theta \) is the angle of projection.

For this specific problem, the vertical component is \( 10 \cdot \frac{1}{2} = 5 \; \text{m/s} \).
Next, we identify the horizontal component with:

• \( u_x = u \cdot \cos(\theta) \)

This results in a value of \( 10 \cdot \frac{\sqrt{3}}{2} = 8.66 \; \text{m/s} \).
Both components play crucial roles: the vertical component affects the upward and downward motion, while the horizontal component determines how far the ball travels along the ground. Together, they provide a comprehensive view of the projectile's movement.

The horizontal distance a projectile travels is determined by its horizontal velocity and the time it remains in flight. In simpler terms, horizontal distance is just how far the projectile can move along the ground while still in the air.
Once we know the initial horizontal velocity component, \( u_x \), we can use it to find the distance covered during the flight using the formula:

• \( d = u_x \cdot t \)
• "d" represents the horizontal distance.

In this scenario, \( u_x = 8.66 \; \text{m/s} \) and \( t = 1 \; \text{second} \) (the time calculated for the ball to return to the roof height).
Thus, the ball travels a horizontal distance of \( 8.66 \) meters. This distance is reached when the ball is at the same vertical position as it started on the roof, effectively completing a segment of its parabolic path.

In physics, equations of motion describe the behavior of a moving object. They are particularly useful in projectile motion problems like this one.
For vertical motion, an equation that relates the height, initial velocity, and time can be expressed as:

• \( h = u_y \cdot t - 0.5 \cdot g \cdot t^2 \)
• where \( h \) is the height, \( g \) is the gravitational acceleration.

Applying this equation helps determine the time when the ball returns to the roof's height: \( 10 = 5 \cdot t - 0.5 \cdot 10 \cdot t^2 \).
Restructuring and solving this quadratic equation yields \( t = 1 \; \text{second} \), meaning it takes this time for the ball to rise and fall back to the initial vertical level.
Understanding and using appropriate equations of motion allow us to predict where and when events happen, crucial for solving projectile motion problems effectively.