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A golf ball is struck with a five iron on level ground. It lands \(92.2 \mathrm{m}\) away \(4.30 \mathrm{s}\) later. What were (a) the direction and (b) the magnitude of the initial velocity?

Short Answer

Expert verified
(a) Direction is approximately 44.74° above the horizontal. (b) Magnitude of initial velocity is approximately 30.14 m/s.

Step by step solution

01

Identify Known Values

We know the horizontal distance traveled by the golf ball is 92.2 m, and the total time of flight is 4.30 s. We need to find the direction (angle) and magnitude of the initial velocity.
02

Establish the Equations of Motion

The horizontal motion can be described by the equation \( x = v_{0x}t \). The vertical motion can be described by \( y = v_{0y}t - \frac{1}{2}gt^2 \), where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Since the ball lands on level ground, \( y = 0 \).
03

Solve for Horizontal Component of Velocity

Using \( x = v_{0x}t \), where \( x = 92.2 \text{ m} \) and \( t = 4.30 \text{ s} \), we can solve for \( v_{0x} \):\[ v_{0x} = \frac{x}{t} = \frac{92.2 \text{ m}}{4.30 \text{ s}} = 21.44 \text{ m/s} \]
04

Solve for Vertical Component of Velocity

Using the vertical motion equation \( y = v_{0y}t - \frac{1}{2}gt^2 \) and knowing \( y = 0 \), we rearrange to find:\[ 0 = v_{0y}t - \frac{1}{2}gt^2 \Rightarrow v_{0y} = \frac{1}{2}gt \]Substituting \( g = 9.8 \text{ m/s}^2 \) and \( t = 4.30 \text{ s} \):\[ v_{0y} = \frac{1}{2} \cdot 9.8 \text{ m/s}^2 \cdot 4.30 \text{ s} = 21.07 \text{ m/s} \]
05

Calculate Magnitude of Initial Velocity

The magnitude of the initial velocity \( v_0 \) can be found using Pythagoras’ theorem:\[ v_0 = \sqrt{v_{0x}^2 + v_{0y}^2} = \sqrt{(21.44)^2 + (21.07)^2} \approx 30.14 \text{ m/s} \]
06

Determine Direction of Initial Velocity

The direction, or angle \( \theta \), of the initial velocity with respect to the horizontal, can be found using:\[ \theta = \tan^{-1} \left( \frac{v_{0y}}{v_{0x}} \right) = \tan^{-1} \left( \frac{21.07}{21.44} \right) \approx 44.74^\circ \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a branch of physics that details the motion of objects without addressing the forces that cause such motion. In projectile motion, like the flight of a golf ball, understanding kinematics is crucial.
Projectile motion involves two components: horizontal and vertical, each moving independently of the other. This dual component setup lets us explore paths that objects take when launched into the air, like balls or arrows.
In kinematics, we utilize the equations of motion to decode these paths. For horizontal motion, things are simpler since there’s no acceleration (ignoring air resistance). For our golf ball, this motion is uniform. However, in the vertical direction, gravity acts as an acceleration of approximately \(9.8\, \mathrm{m/s^2}\). This acceleration affects the vertical ride of the projectile, altering its height and descent.
Because of this independence, we can separate the kinematic equations for horizontal and vertical analysis, looking individually at velocity, position, and time factors. Thus, kinematics provides a detailed picture of how the golf ball moves through its trajectory, essential for accurately determining properties like initial velocity.
Initial Velocity
The concept of initial velocity is central to understanding projectile motion, as it is the velocity at which the projectile is launched. It dictates how far and fast the projectile will travel.
The initial velocity has two components:
  • Horizontal Velocity (\( v_{0x} \)
  • Vertical Velocity (\( v_{0y} \)
In this exercise, we derive these components separately.
For the horizontal component, we use the relation \( x = v_{0x} t \), deriving\( v_{0x} = 21.44\, \mathrm{m/s} \).For the vertical, using the rearranged form\( y = v_{0y} t - \frac{1}{2} g t^2 \), we find\( v_{0y} = 21.07\, \mathrm{m/s} \).
These separate calculations illustrate how the golf ball begins its journey through both horizontal space and gravitational pull.
Upon combining these components, we can identify the initial velocity's magnitude using the Pythagorean theorem, resulting in a value of approximately\( 30.14\, \mathrm{m/s} \).
The angle of projection, or the direction concerning the horizontal, enhances our comprehension of how this initial velocity guides a projectile in its path.
Equations of Motion
The equations of motion are mathematical relationships that describe a moving object's behavior over time. They are essential for solving problems in projectile motion.
In the context of our exercise, two equations are principally employed:
  • The horizontal equation: \( x = v_{0x}t \)
  • The vertical equation: \( y = v_{0y}t - \frac{1}{2}gt^{2} \)
These equations help us break down the motion into manageable parts and decipher unknowns like initial velocity and time of flight.
With the horizontal equation, we easily determine the horizontal velocity, given that horizontal motion remains constant in the absence of air resistance. Conversely, the vertical equation accounts for acceleration due to gravity, which is crucial as it asserts that the upward motion will eventually slow and reverse.
The interactions described by these equations let us solve for both the magnitude and angle of a projectile's initial velocity, pivotal for predicting its range and behavior upon launch.
Each piece of the puzzle—horizontal and vertical—comes from these core kinematic equations, forming a cohesive picture of the object's journey through space and time.

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