Problem 32 - Fundamentals of Physics

Question

You throw a ball toward a wall at speed \(25.0 \mathrm{~m} / \mathrm{s}\) and at angle \(\theta_{0}=40.0^{\circ}\) above the horizontal (Fig. 4-35). The wall is distance \(d=\) \(22.0 \mathrm{~m}\) from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory?

Step-by-Step Solution

Answer

(a) 8.58 m above release. (b) 19.15 m/s; (c) 4.75 m/s. (d) No, not highest yet.

Step 1: Decompose Initial Velocity

We need to find the horizontal and vertical components of the initial velocity. The horizontal component \( v_{0x} \) is given by \( v_{0x} = v_0 \cos \theta_0 \). The vertical component \( v_{0y} \) is given by \( v_{0y} = v_0 \sin \theta_0 \). Here, \( v_0 = 25.0 \, \text{m/s} \) and \( \theta_0 = 40.0^{\circ} \). Calculating these, we get:\[ v_{0x} = 25.0 \, \cos 40.0^{\circ} \approx 19.15 \, \text{m/s}\]\[ v_{0y} = 25.0 \, \sin 40.0^{\circ} \approx 16.07 \, \text{m/s}\]

Step 2: Calculate Time to Reach the Wall

Use the horizontal motion equation to calculate the time \( t \) it takes for the ball to reach the wall: \( d = v_{0x} \cdot t \). Solving for \( t \), we find:\[ t = \frac{d}{v_{0x}} = \frac{22.0 \, \text{m}}{19.15 \, \text{m/s}} \approx 1.15 \, \text{s} \]

Step 3: Calculate Vertical Position at the Wall

We use the vertical motion equation to determine how far above the release point the ball is when it hits the wall:\[ y(t) = v_{0y} \cdot t - \frac{1}{2} g t^2 \]Substituting the known values:\[ y(1.15) = 16.07 \, \text{m/s} \cdot 1.15 \, \text{s} - \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 \cdot (1.15 \, \text{s})^2 \approx 8.58 \, \text{m} \]

Step 4: Calculate Horizontal Velocity at Wall

The horizontal velocity remains constant throughout the flight because there is no horizontal acceleration. Thus, \( v_x = v_{0x} = 19.15 \, \text{m/s} \).

Step 5: Calculate Vertical Velocity at Wall

The vertical velocity \( v_y \) can be calculated using the equation \( v_y = v_{0y} - g t \). Substituting the values:\[ v_y = 16.07 \, \text{m/s} - 9.8 \, \text{m/s}^2 \cdot 1.15 \, \text{s} \approx 4.75 \, \text{m/s} \]

Step 6: Evaluate If Ball Has Passed the Highest Point

The ball reaches its highest point when its vertical velocity is zero, \( v_y = 0 \). Since the vertical velocity \( v_y \) at the wall is positive (4.75 m/s), the ball has not yet reached its highest point.

Key Concepts

Initial Velocity Components

Understanding the initial velocity components is crucial in projectile motion problems. When you launch any object, like a ball, at a specific angle, it has an initial speed. This speed can be decomposed into two components: horizontal and vertical.
This is particularly important because each component will influence the object's trajectory independently.

• Horizontal Component \( (v_{0x}) \): This is found using the formula \( v_{0x} = v_0 \cos \theta_0 \). It remains constant throughout the flight, as there is no horizontal acceleration (ignoring air resistance).
• Vertical Component \( (v_{0y}) \): This is calculated using \( v_{0y} = v_0 \sin \theta_0 \). The vertical component will change due to the effect of gravity.

For the exercise given, with an angle on the horizontal being \( \theta_0=40.0^{\circ} \) and initial speed \( 25.0 \, \text{m/s} \), the horizontal velocity was calculated to be approximately \( 19.15 \, \text{m/s} \), while the vertical velocity was \( 16.07 \, \text{m/s} \). These components help us track how far and how high the projectile will travel.

Trajectory Analysis

The trajectory of an object in projectile motion describes the path it follows through space. Analyzing this path can tell you a lot about how far and high the object will go.
Projectile motion is affected by both the vertical and horizontal components of motion. Understanding the time it takes to reach a point is critical in trajectory analysis.

• Horizontal Motion: Governs the distance traveled. Since there is no horizontal acceleration, distance is calculated as: \( d = v_{0x} \cdot t \). Using this relation, time \( t \) can be determined.
• Vertical Motion: The height at any point is given by \( y(t) = v_{0y} \cdot t - \frac{1}{2}gt^2 \). This equation accounts for both the initial upward velocity and the acceleration due to gravity \( g \).

In the given exercise, determining the time to reach the wall was necessary to further find the vertical position of the ball at that time. The time was calculated as about \( 1.15 \, \text{s} \), resulting in a height of around \( 8.58 \, \text{m} \) above the release point.

Vertical Motion Equations

Vertical motion in projectile problems is complex due to gravity's influence, which causes the vertical velocity to change over time. These changes are described by key equations.

• Vertical Position: Given by \( y(t) = v_{0y} \cdot t - \frac{1}{2}gt^2 \), this formula calculates the vertical position relative to its starting point. This accounts for initial upward velocity and gravitational pull.
• Vertical Velocity: The velocity at any point is calculated with \( v_y = v_{0y} - gt \). It decreases over time because of gravity at \( 9.8 \, \text{m/s}^2 \).

In our exercise, to find the speed at which the ball hits the wall, you use the vertical velocity calculation. Here, after \( 1.15 \, \text{s} \), we found that the vertical velocity was approximately \( 4.75 \, \text{m/s} \), showing the ball still moving upwards, indicating it hasn’t reached its peak height yet.