/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 At the 18 th green of the U.S. O... [FREE SOLUTION] | 91Ó°ÊÓ

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At the 18 th green of the U.S. Open you need to make a \(20.5-f t\) putt to win the tournament. When you hit the ball, giving it an initial speed of \(1.57 \mathrm{m} / \mathrm{s},\) it stops \(6.00 \mathrm{ft}\) short of the hole. (a) Assuming the deceleration caused by the grass is constant, what should the initial speed have been to just make the putt? (b) What initial speed do you need to make the remaining \(6.00-\mathrm{ft}\) putt?

Short Answer

Expert verified
(a) 1.86 m/s (b) 1.01 m/s

Step by step solution

01

Convert Distances to Meters

The initial putt distance is 20.5 feet. The ball stops 6 feet short, so it actually travels 14.5 feet. Convert feet to meters (1 foot = 0.3048 meters). Calculate: - Total intended distance: \[ 20.5 \times 0.3048 = 6.2484 \text{ m} \]- Actual travel distance: \[ 14.5 \times 0.3048 = 4.4196 \text{ m} \] - Remaining distance: \[ 6.0 \times 0.3048 = 1.8288 \text{ m} \]
02

Calculate the Deceleration

Use the formula of motion with constant acceleration: \[ v^2 = u^2 + 2as \]Where:- \(v\) is the final speed (0 m/s, since the ball stops)- \(u\) is the initial speed (1.57 m/s)- \(a\) is the acceleration- \(s\) is the distance traveled (4.4196 m)Rearrange to find \(a\):\[ a = \frac{v^2 - u^2}{2s} = \frac{0 - (1.57)^2}{2 \times 4.4196} \]Calculate \(a\):\[ a = \frac{-2.4649}{8.8392} = -0.2786 \text{ m/s}^2 \]
03

Calculate Required Initial Speed for Full Putt

Use the same kinematic formula for a putt just reaching the hole:\[ v^2 = u^2 + 2as \]Now \(s\) is the full intended distance (6.2484 m).Solve for \(u\):\[ 0 = u^2 + 2(-0.2786)(6.2484) \]\[ u^2 = 2 \times 0.2786 \times 6.2484 \]\[ u^2 = 3.4753 \]\[ u = \sqrt{3.4753} \approx 1.863 \text{ m/s} \]
04

Calculate Required Initial Speed for Remaining Putt

Apply the same formula for the remaining 6 ft (1.8288 m):\[ v^2 = u^2 + 2as \]Solve for \(u\):\[ 0 = u^2 + 2(-0.2786)(1.8288) \]\[ u^2 = 2 \times 0.2786 \times 1.8288 \]\[ u^2 = 1.0183 \]\[ u = \sqrt{1.0183} \approx 1.009 \text{ m/s} \]

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Key Concepts

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

Deceleration
Deceleration occurs when an object's speed decreases with time. This is essentially the opposite of acceleration. When hitting a golf ball on a putting green, the surface of the grass creates resistance that slows down the ball, causing it to decelerate. In physics, deceleration is considered a type of acceleration because it involves a change in velocity. A key aspect of this exercise is that the deceleration is constant, meaning the rate at which the ball's speed decreases remains steady as it travels. To find deceleration, we use the kinematic equation:
  • Final velocity ( v ) = 0 since the ball stops.
  • Initial velocity ( u ) is 1.57 m/s.
  • Distance ( s ) the ball travels is 4.4196 meters.
By determining the rate at which the ball slows, we better understand what initial speed is necessary to cover specific distances.
Initial Speed
Initial speed, or initial velocity, refers to how fast an object is moving when it starts. In our exercise, the initial speed of the golf ball was 1.57 meters per second. This speed, however, was insufficient to cover the entire 20.5-foot distance as it stopped 6 feet short, only managing 14.5 feet. Adjusting the initial speed is crucial, especially in sports where precision is required. It determines whether or not the ball will reach its desired destination, overcoming resistance such as friction from the grass on a putting green. Calculating the precise initial speed required to reach the hole fully means we can address any needed changes based on the deceleration the grass causes.
Constant Acceleration
Constant acceleration implies that an object's rate of change for speed stays the same over time. This is particularly useful when calculating motion, as it simplifies the equations we use. In this problem, although the term used is deceleration, it fits under constant acceleration as the rate remains the same.The equation \[ a = \frac{v^2 - u^2}{2s} \] allows us to find the acceleration (or deceleration) easily. With deceleration known as -0.2786 m/s² after rearranging the equation, it provides a straightforward way to predict how the ball's speed changes over the distance traveled. This concept allows for consistent results and simple calculations.
Kinematic Equations
Kinematic equations are a set of equations that describe motion without regard to its causes. They are particularly valuable for solving problems involving objects moving in a straight line with constant acceleration. In this problem, the primary kinematic equation used is:\[ v^2 = u^2 + 2as \]These equations help determine the various unknowns in motion problems.
  • For calculating deceleration: helps find how quickly a ball slows down.
  • For finding initial speed: calculates based on distance to travel and deceleration.
Multiple kinematic equations exist, with different setups for specific variables related to time, speed, acceleration, and distance, enhancing our understanding and ability to predict motion outcomes accurately.

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