Question

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.480 s. Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. (d) Draw \(x-t, y-t, v_x-t\), and \(v_y-t\) graphs for the motion.

Short answer

The table is approximately 1.13 m high, the horizontal distance is 0.528 m, velocities are 1.10 m/s horizontal and 4.709 m/s vertical, with a magnitude of 4.829 m/s at 76.8° below horizontal.

Step-by-step solution

Calculate the height of the tabletop (a)

To find the height of the table, we use the formula for vertical motion: \( h = \frac{1}{2} g t^2 \). The acceleration due to gravity \( g \) is approximately \( 9.81 \text{ m/s}^2 \), and the time \( t \) is \( 0.480 \text{ s} \). So we have: \[ h = \frac{1}{2} \times 9.81 \times (0.480)^2 \approx 1.13 \text{ m} \] Therefore, the height of the tabletop is approximately 1.13 meters.

Calculate the horizontal distance traveled (b)

The horizontal distance can be found using the formula \( d = v_0 \cdot t \), where \( v_0 \) is the initial horizontal speed (1.10 m/s) and \( t \) is the time taken (0.480 s): \[ d = 1.10 \times 0.480 \approx 0.528 \text{ m} \] Thus, the horizontal distance from the table to where the book strikes the floor is approximately 0.528 meters.

Calculate the horizontal and vertical velocity components (c1)

\( v_x \) remains constant because there is no horizontal acceleration, hence \( v_x = 1.10 \text{ m/s} \). For the vertical component \( v_y \), use \( v_y = g \times t \): \[ v_y = 9.81 \times 0.480 \approx 4.709 \text{ m/s} \] The horizontal and vertical components of the velocity just before the book reaches the floor are 1.10 m/s and approximately 4.709 m/s, respectively.

Find the magnitude and direction of the total velocity (c2)

The magnitude of the velocity vector can be found using the Pythagorean theorem: \[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{1.10^2 + 4.709^2} \approx 4.829 \text{ m/s} \] The direction (angle θ below the horizontal) can be found using: \[ \theta = \arctan\left(\frac{v_y}{v_x}\right) = \arctan\left(\frac{4.709}{1.10}\right) \approx 76.8^\circ \] Thus, the magnitude of the velocity is about 4.829 m/s, in a direction approximately 76.8 degrees below the horizontal.

Sketch the graphs (d)

For the path of the book, the graphs are as follows:- x-t graph: A straight line with a constant positive slope of 1.10 m/s, indicating uniform motion.- y-t graph: A parabolic curve starting at the origin (0,0) and steepening, showing increasing velocity.- \(v_x-t\) graph: A flat line at 1.10 m/s, indicating constant horizontal velocity.- \(v_y-t\) graph: A straight line with a positive slope, starting from zero, indicating constant vertical acceleration.

Key concepts

Kinematics

In physics, kinematics is the study of motion without considering the forces that cause it. It focuses on the movement of objects, described in terms of displacement, velocity, and acceleration. When a book falls off a tabletop, we are interested in how it moves over time, paying close attention to the changes in its motion. Kinematics allows us to predict where and how fast the book will be at any given time. By using equations specific to both vertical and horizontal motion, we gain detailed insights into the trajectory and speed of the book. This exploration into kinematics gives us the tools to better understand complex motion without diving into the forces involved.

Horizontal Motion

Horizontal motion refers to movement along a straight path parallel to the ground. In our scenario, the book slides off the table with an initial speed of 1.10 m/s. Importantly, since there is no horizontal force acting on it (as we ignore air resistance), the book's horizontal velocity remains constant throughout its fall. This consistency simplifies calculations, as we can simply multiply velocity by time to find the horizontal distance traveled:

• Initial horizontal speed, \(v_x = 1.10 \text{ m/s}\)
• Time of fall, \(t = 0.480 \text{ s}\)
• Horizontal distance, \(d = v_x \cdot t = 0.528 \text{ m}\)

This formula applies when horizontal acceleration is zero, a key assumption in many projectile motion problems.

Vertical Motion

Vertical motion adds a layer of complexity due to gravity, which constantly accelerates the book downward. This acceleration affects how quickly the book reaches the ground. The height from which the book falls can be calculated using the equation:

• Height, \(h = \frac{1}{2} g t^2\)
• Using \(g = 9.81 \text{ m/s}^2\) and \(t = 0.480 \text{ s}\)
• Gives \(h \approx 1.13 \text{ m}\)

Furthermore, the vertical velocity just before impact is determined by multiplying gravity by time: \(v_y = g \cdot t\). This aspect is crucial in understanding how quickly the book is moving downward as it hits the floor, enhancing our comprehension of motion affected by gravity.

Velocity Components

Every motion in physics can be broken down into components. In projectile motion, these are typically horizontal and vertical. Knowing the components allows us to use basic trigonometry for analysis and predictions:

• Horizontal component, \(v_x = 1.10 \text{ m/s}\)
• Vertical component, \(v_y = 9.81 \times 0.480 = 4.709 \text{ m/s}\)

The magnitude of the resultant velocity vector is calculated by combining these components using the Pythagorean theorem:\[ v = \sqrt{v_x^2 + v_y^2} \approx 4.829 \text{ m/s} \]To find the direction, you calculate the angle \(\theta\) below the horizontal using:\[ \theta = \arctan\left(\frac{v_y}{v_x}\right) \approx 76.8^\circ \]This analysis helps understand not only how fast the book is moving but also the angle at which it descends, crucial for precise predictions of its landing position.