Problem 8 - College Physics

Question

A physics book slides off a horizontal tabletop with a speed of 1.10 \(\mathrm{m} / \mathrm{s}\) . It strikes the floor in 0.350 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, and (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.

Step-by-Step Solution

Answer

a) Height is 0.602 m. b) Horizontal distance is 0.385 m. c) Velocity components are 1.10 m/s and 3.434 m/s; magnitude is 3.61 m/s, direction is 72.7° below horizontal.

Step 1: Determine the vertical distance (height)

Using the equation of motion for free fall, calculate the height the book falls. The formula for free fall is;\[ h = \frac{1}{2}gt^2 \]where:- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)- \( t = 0.350 \, \text{s} \) (time taken to fall)Substitute the values:\[ h = \frac{1}{2} \times 9.81 \times (0.350)^2 \]\[ h = 0.602 \text{ meters} \]

Step 2: Calculate the horizontal distance

Use the horizontal component of the motion to find the distance (\(x\)) the book travels horizontally. The formula is:\[ x = v_x \cdot t \]where:- \( v_x = 1.10 \, \text{m/s} \) (horizontal speed)Substitute the values:\[ x = 1.10 \times 0.350 \]\[ x = 0.385 \, \text{meters} \]

Step 3: Find the horizontal and vertical components of velocity

Calculate the velocity components just before the book hits the ground.- Horizontal velocity is constant: \( v_x = 1.10 \, \text{m/s} \)- Vertical velocity can be calculated as:\[ v_y = gt \]Substitute the values:\[ v_y = 9.81 \times 0.350 \]\[ v_y = 3.434 \, \text{m/s} \]

Step 4: Determine the magnitude of the velocity

Calculate the magnitude of the velocity using Pythagorean theorem:\[ v = \sqrt{v_x^2 + v_y^2} \]Substitute the calculated components:\[ v = \sqrt{(1.10)^2 + (3.434)^2} \]\[ v \approx 3.61 \, \text{m/s} \]

Step 5: Calculate the direction of the velocity

The direction (angle \( \theta \)) with respect to the horizontal can be found using the tangent function:\[ \theta = \arctan \left( \frac{v_y}{v_x} \right) \]Substitute the values:\[ \theta = \arctan \left( \frac{3.434}{1.10} \right) \]\[ \theta \approx 72.7^\circ \] below the horizontal.

Key Concepts

Free Fall Motion

In physics, free fall motion describes the movement of an object exclusively under the influence of gravity. When analyzing such motion, we often dismiss factors like air resistance to simplify calculations. The acceleration due to gravity on Earth is approximately 9.81 m/s². This constant force pulls objects downward.

• In our example, a book falls from a tabletop, moving straight down.
• The formula for calculating the distance fallen is: \[ h = \frac{1}{2}gt^2 \]

Knowing the time taken by the book to hit the ground (0.350 seconds), we can plug this into the formula to find the height: \[ h = \frac{1}{2} \times 9.81 \times (0.350)^2 = 0.602 \text{ meters} \]
This result shows how free fall tightly links to both time and gravity's constant pull.

Horizontal Distance Calculation

While the book is falling, it also travels horizontally. This part of its motion isn't influenced by gravity, and so it maintains a consistent horizontal velocity throughout. In this situation:

• The horizontal speed of the book is constant at 1.10 m/s due to the flat surface of the table.
• The formula employed to find the horizontal distance is: \[ x = v_x \cdot t \]

With given values: \[ x = 1.10 \times 0.350 = 0.385 \text{ meters} \]
This calculation gives the horizontal range the book covers when it falls from the table to the ground. Understanding constant horizontal speed is critical in analyzing projectile motion.

Velocity Components

Projectile motion can be decomposed into horizontal and vertical components, allowing easier analysis. Consistent horizontal velocity (\(v_x\)) is unaffected by gravity, while vertical velocity (\(v_y\)) steadily increases as the book falls.

• Horizontal velocity remains at 1.10 m/s.
• Vertical velocity at impact can be calculated with: \[v_y = gt\]

Given vertical acceleration, calculate \(v_y\): \[v_y = 9.81 \times 0.350 = 3.434 \text{ m/s} \]
The two velocities create a vector, influencing how the book moves right before it touches the ground. This approach improves understanding of the dynamics in projectile motion.

Angle of Velocity

The angle of velocity determines the direction of an object's resultant velocity vector. Calculating this angle requires knowing both horizontal and vertical velocity components. The magnitude of the velocity is calculated first,\[v = \sqrt{v_x^2 + v_y^2} \]. For the book, the values are:
\[v = \sqrt{(1.10)^2 + (3.434)^2} \approx 3.61 \text{ m/s} \].
To find the angle \( \theta \) relative to the horizontal, use the arctangent function:

• \[\theta = \arctan \left( \frac{v_y}{v_x} \right)\]

Substitute in the known velocities:\( \theta = \arctan \left( \frac{3.434}{1.10} \right) \approx 72.7^\circ \), angled below the horizontal.
This angle describes the trajectory of the object before contact, revealing projectile direction.