Problem 2 - College Physics

Question

A golf ball is hit with an initial speed of \(35 \mathrm{~m} / \mathrm{s}\) at an angle less than \(45^{\circ}\) above the horizontal. (a) The horizontal velocity component is (1) greater than, (2) equal to, (3) less than the vertical velocity component. Why? (b) If the ball is hit at an angle of \(37^{\circ},\) what are the initial horizontal and vertical velocity components?

Step-by-Step Solution

Answer

(a) (1) Greater than. (b) Horizontal: 27.951 m/s, Vertical: 21.063 m/s.

Step 1: Understanding the Problem

The problem asks about the relationship between the horizontal and vertical components of a golf ball's velocity when hit at an angle less than 45°, specifically 37°. We need to compare these components and calculate them based on the given angle and initial speed.

Step 2: Analyzing the Angle

Angles below 45° result in a horizontal velocity component that is greater than the vertical component. This is because as the angle decreases, the horizontal component (cosine component) increases relative to the vertical component (sine component). Hence, when the angle is less than 45°, the horizontal velocity is greater than the vertical velocity.

Step 3: Calculating Horizontal Velocity

To find the horizontal velocity component, use the formula: \( v_{x} = v \, \cos(\theta) \). Here, \( v = 35 \, \mathrm{m/s} \) and \( \theta = 37^{\circ} \). Thus, \( v_{x} = 35 \, \cos(37^{\circ}) \). Using a calculator, \( \cos(37^{\circ}) \approx 0.7986 \), so \( v_{x} \approx 35 \times 0.7986 \approx 27.951 \, \mathrm{m/s} \).

Step 4: Calculating Vertical Velocity

To find the vertical velocity component, use the formula: \( v_{y} = v \, \sin(\theta) \). Here, \( v = 35 \, \mathrm{m/s} \) and \( \theta = 37^{\circ} \). Thus, \( v_{y} = 35 \, \sin(37^{\circ}) \). Using a calculator, \( \sin(37^{\circ}) \approx 0.6018 \), so \( v_{y} \approx 35 \times 0.6018 \approx 21.063 \, \mathrm{m/s} \).

Step 5: Summarizing Results

For a launch angle of 37°, the horizontal velocity component \( v_{x} \) is approximately 27.951 m/s and the vertical velocity component \( v_{y} \) is approximately 21.063 m/s. The horizontal component is greater than the vertical component, as expected based on our angle analysis.

Key Concepts

Horizontal Velocity

Horizontal velocity is the component of a projectile's velocity that is parallel to the Earth's surface. In projectile motion, the horizontal velocity remains constant if we neglect air resistance. To calculate the horizontal velocity, we use the formula:\[ v_{x} = v \cos(\theta) \]where:

• \(v\) is the initial velocity of the projectile.
• \(\theta\) is the angle of projection above the horizontal.

In the case of the golf ball hit at an angle of 37° with an initial speed of 35 m/s, the horizontal component of velocity \(v_{x}\) can be calculated by substituting the given values into the formula, resulting in approximately 27.951 m/s.Understanding the calculation helps you predict the behavior of the projectile's motion, essential for solving various physics problems.

Vertical Velocity

Vertical velocity is the component of velocity that acts perpendicular to the ground. Unlike horizontal velocity, vertical velocity in projectile motion changes due to the influence of gravity, which causes a uniform acceleration downwards. The formula to find the vertical velocity component is:\[ v_{y} = v \sin(\theta) \]where:

• \(v\) is the initial speed.
• \(\theta\) is the angle of launch.

For the golf ball with an initial speed of 35 m/s at 37°, the vertical velocity \(v_{y}\) is calculated to be approximately 21.063 m/s. Gravity will affect this component, slowing it down as the projectile ascends and speeding it up as it descends.Grasping the concept of vertical velocity is crucial to understanding how high and how long the projectile will travel upwards before starting to descend.

Velocity Components

Projectile motion involves the separation of a projectile's velocity into two components: horizontal and vertical. These components are independent of each other, allowing us to analyze them separately. The initial velocity \(v\) can be divided into horizontal \(v_{x}\) and vertical \(v_{y}\) components using trigonometric functions:

• Horizontal velocity component: \( v_{x} = v \cos(\theta) \)
• Vertical velocity component: \( v_{y} = v \sin(\theta) \)

These two components collectively define the projectile's motion path. For angles less than 45°, the horizontal component is greater than the vertical component. This is because the cosine of the angle is larger than the sine.Understanding these components helps you break down complex projectile motion problems, making it easier to predict and calculate the trajectory and final position of the object.