91Ó°ÊÓ

In projectile motion like the situation of a book sliding off a tabletop, horizontal motion is described as movement along the horizontal axis without any acceleration (if air resistance is ignored). From the original exercise, the initial horizontal velocity (\( v_{i_x} \)) of the book is given as 1.10 m/s. This value remains constant throughout the motion because there are no external forces acting horizontally to change it.
Without acceleration, we can use the simple formula for horizontal distance:

• \( x = v_{i_x} \cdot t \)

Where \( x \) is the horizontal distance and \( t \) is the time taken.
This means that the book travels 0.385 meters horizontally from the edge of the table to the floor, calculated by multiplying the horizontal velocity by the time of 0.350 seconds.

Vertical motion during projectile motion is influenced by gravity, which provides a constant acceleration downwards. In the given problem, the initial vertical velocity (\( v_{i_y} \)) at the moment the book starts falling is 0. This is because the book simply slides off a horizontal surface without any initial vertical push.
The formula used to determine the vertical displacement (\( y \)) is:

• \( y = \frac{1}{2} g t^2 \)

Here, \( g \) represents the gravitational acceleration which is approximately 9.8 m/s². Plugging in the time of 0.350 seconds results in a vertical displacement or height of 0.602 meters. This shows how far the book falls vertically as it travels horizontally.

A crucial aspect of understanding projectile motion is breaking the velocity into horizontal and vertical components. Each component acts independently.

• The horizontal component \( v_x \) remains 1.10 m/s from the beginning to the end since there is no horizontal acceleration.
• The vertical component \( v_y \) changes over time due to gravitational acceleration, calculated using:
  • \( v_y = g \cdot t \)
  With gravitational effect and a time of 0.350 seconds, \( v_y \) becomes 3.43 m/s downward just before collision with the floor.
  

Combining these components allows for calculating the overall magnitude and direction of the velocity vector via:

• \( v = \sqrt{v_x^2 + v_y^2} \)
• Direction: \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \)

The speed of 3.59 m/s and direction of approximately 72.9 degrees below the horizontal reflect its final velocity state.

Displacement in projectile motion takes into account both horizontal and vertical movements. In the problem of the textbook's journey off the table, displacement involves calculating the distance it travels in both directions, combining to define its overall trajectory.
The horizontal displacement is the 0.385 meters it travels from the edge of the table, while the vertical displacement is equivalent to the height of the table, 0.602 meters.

• This trajectory can be visualized through graphs:
• The position vs. time graph for horizontal motion (x-t graph) appears as a straight line, showing constant horizontal speed.
• The position vs. time graph for vertical motion (y-t graph) appears as a curve due to the acceleration downward from gravity.
Drawing these graphs helps visualize how horizontal and vertical displacements combine during the motion to form a parabolic trajectory, typical of projectiles like our textbook.