91Ó°ÊÓ

In projectile motion, velocity components are crucial. These components refer to the velocity of an object which can be broken into horizontal and vertical parts. The horizontal component, often denoted as \( v_x \), signifies how fast the object is moving along a horizontal path.
Meanwhile, the vertical component, denoted as \( v_y \), indicates how quickly the object is moving up or down.

The beauty of analyzing velocity in this way is in its simplicity. Instead of considering the object's complex trajectory, we separate it into easier-to-understand parts. In the given exercise, the ball has a horizontal velocity of \(+10.0 \text{ m/s}\) and a downward vertical velocity of \(15.0 \text{ m/s}\).

• Horizontal components are always constant in the absence of air resistance, as no forces act horizontally.
• Vertical components change due to gravitational forces, which can speed up or slow down the object.

This separation allows you to grasp intricate motion by focusing on straightforward, individual paths.

The angle of trajectory is the angle at which the object travels with respect to the horizontal plane. It's what gives projectile motion its signature curve. To find this angle, you need the tangent function, which compares the vertical and horizontal velocities.
The formula used is:

• \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \)

This calculation leverages the inverse tangent function, \( \tan^{-1} \), to find the angle \( \theta \) from the ratio of vertical to horizontal components. In our case, substituting \( v_y = 15.0 \text{ m/s} \) and \( v_x = 10.0 \text{ m/s} \) yields:
\[ \theta = \tan^{-1}\left(\frac{15.0}{10.0}\right) = \tan^{-1}(1.5) \approx 56.31^\circ \]

This angle tells us how steeply the ball is leaving the roof, providing insight into both its initial direction and the expected path.

Resultant speed in projectile motion is the overall speed at which an object travels, factoring in both horizontal and vertical components. It is not just a simple sum of these components, but rather the hypotenuse of a right triangle formed by them. You calculate this using the Pythagorean theorem.
The equation to find the resultant speed \( v \) is:

• \( v = \sqrt{v_x^2 + v_y^2} \)

Substituting the given values, \( v_x = 10.0 \text{ m/s} \) and \( v_y = 15.0 \text{ m/s} \), we calculate:
\[ v = \sqrt{(10.0)^2 + (15.0)^2} = \sqrt{100 + 225} = \sqrt{325} \approx 18.03 \text{ m/s} \]
This resultant speed represents the actual velocity at which the ball leaves the rooftop. It captures the full impact of both directional motions and gives a clearer picture of the ball's projected motion into space.