Problem 79 - College Physics

Question

IE A 0.455 -kg soccer ball is kicked off level ground at an angle of \(40^{\circ}\) with an initial speed of \(30.0 \mathrm{~m} / \mathrm{s}\). Neglecting air resistance, (a) at its maximum height off the ground, its kinetic energy will be (1) less than its value at launch, but not zero, (2) more than its value at launch, (3) zero. Explain. (b) Determine its kinetic energy when it is at its maximum height above the ground and compare it to the kinetic energy at launch. [Hint: What is its velocity at the top of its arc? Review projectile motion if necessary.]

Step-by-Step Solution

Answer

(a) Kinetic energy is less than at launch, but not zero. (b) Maximum height KE = 119.5 J; less than launch KE (204.75 J).

Step 1: Analyze Initial Conditions

First, note the information provided: the mass of the soccer ball is 0.455 kg, the launch angle is \( 40^{\circ} \), and the initial speed is 30.0 m/s. Use this information to determine the initial kinetic energy using the formula \( KE = \frac{1}{2}mv^2 \). For the initial kinetic energy at launch, plug in \( m = 0.455 \) kg and \( v_0 = 30.0 \) m/s.

Step 2: Calculate Initial Kinetic Energy

Using the formula \( KE = \frac{1}{2}mv^2 \), the initial kinetic energy is \( KE = \frac{1}{2} \times 0.455 \times (30.0)^2 \). This gives \( KE = 204.75 \) J.

Step 3: Break Down Projectile Motion

Divide the initial speed into horizontal and vertical components. The horizontal component \( v_{0x} = v_0 \cos(40^{\circ}) \) and the vertical component \( v_{0y} = v_0 \sin(40^{\circ}) \). This gives \( v_{0x} = 30.0 \cos(40^{\circ}) \) m/s and \( v_{0y} = 30.0 \sin(40^{\circ}) \) m/s. Calculate each component.

Step 4: Calculate Horizontal Velocity Component

Calculate the horizontal velocity component: \( v_{0x} = 30.0 \cos(40^{\circ}) \). This results in \( v_{0x} \approx 22.98 \) m/s.

Step 5: Determine Conditions at Maximum Height

At maximum height, the vertical velocity component is zero (\( v_y = 0 \)), but the horizontal component remains constant. Therefore, the velocity at maximum height is entirely horizontal and equals \( v_{0x} \).

Step 6: Calculate Kinetic Energy at Maximum Height

The kinetic energy at maximum height is due only to the horizontal motion: \( KE_{max} = \frac{1}{2}mv_{0x}^2 \). Substitute \( m = 0.455 \) kg and \( v_{0x} = 22.98 \) m/s: \( KE_{max} = \frac{1}{2} \times 0.455 \times (22.98)^2 \approx 119.5 \) J.

Step 7: Compare Kinetic Energies

Compare \( KE_{max} = 119.5 \) J to the initial kinetic energy \( 204.75 \) J. Since \( 119.5 \) J is less than \( 204.75 \) J, the kinetic energy at maximum height is less than at launch, although not zero.

Key Concepts

Kinetic Energy

When it comes to projectile motion, kinetic energy is an essential concept that helps us understand how an object's motion is described in terms of energy. Kinetic energy is the energy an object possesses due to its motion, and it is calculated using the formula \[ KE = \frac{1}{2}mv^2 \]where:

• \( m \) is the mass of the object
• \( v \) is the velocity of the object

In the exercise, we started by calculating the initial kinetic energy of the soccer ball. With a mass of 0.455 kg and an initial velocity of 30 m/s, the kinetic energy at launch is determined to be approximately 204.75 J. At maximum height, although vertical motion ceases, the horizontal kinetic energy remains, resulting in a kinetic energy of about 119.5 J. This informs us that while kinetic energy decreases as the soccer ball reaches its peak, it is not zero.

Horizontal and Vertical Components

Understanding the breakdown of the velocity into horizontal and vertical components is crucial for analyzing projectile motion. The initial velocity of the soccer ball at an angle gets divided into two separate vectors:

• Horizontal Component \( v_{0x} \): This is calculated using \( v\cos(\theta) \), where \( \theta \) is the launch angle (40° in this case). Here, the component works out to approximately 22.98 m/s.
• Vertical Component \( v_{0y} \): This relies on \( v\sin(\theta) \) and, for our exercise, is determined with a value that changes with respect to time.

These components help in predicting how far and how high the ball will travel. Importantly, while the vertical component reaches zero at maximum height (since gravity decelerates it), the horizontal component remains constant, unaffected by gravity when neglecting air resistance.

Maximum Height Analysis

In projectile motion, understanding what happens at the maximum height provides insights into the object's trajectory. When the soccer ball reaches its maximum height, the vertical component of its velocity becomes zero. This is a vital moment because:

• The vertical kinetic energy is completely lost, as gravity has decelerated all upward motion.
• The ball's kinetic energy now entirely consists of horizontal motion.

To analyze the kinetic energy at this peak, consider only the horizontal velocity, which remains unchanged at approximately 22.98 m/s. Many students are surprised to learn that even at the highest point, a portion of kinetic energy persists. As calculated, the energy at maximum height becomes 119.5 J, which corroborates the idea that the kinetic energy is less than at launch but certainly not zero. This peak analysis, while conceptually intricate, highlights how energy conservation and transfer occur within projectile motion.