/*! 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 73 MMH A golfer, standing on a fair... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 73

MMH A golfer, standing on a fairway, hits a shot to a green that is elevated 5.50 m above the point where she is standing. If the ball leaves her club with a velocity of 46.0 m/s at an angle of 35.0 above the ground, find the time that the ball is in the air before it hits the green.

Short Answer

Expert verified
The ball is in the air for approximately 2.35 seconds.

Step by step solution

01

Break Down Initial Velocity

The initial velocity of the golf ball is given as 46.0 m/s at an angle of 35.0 degrees above the ground. We need to resolve this velocity into horizontal and vertical components using trigonometry. The horizontal component is given by \[ v_{x} = v \cos \theta = 46.0 \times \cos(35.0) \]And the vertical component is given by \[ v_{y} = v \sin \theta = 46.0 \times \sin(35.0) \]
02

Calculate Vertical Displacement

The vertical displacement \( y \) is the change in height, which is given as 5.50 m (upwards). Thus, we have \[ y = 5.50 \text{ m} \]
03

Apply Kinematic Equation

Use the kinematic equation for vertical motion to find the time of flight. The equation is \[ y = v_{y} t - \frac{1}{2}gt^2 \]where \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity, acting downwards.
04

Set Up Quadratic Equation

Substitute the known values into the kinematic equation: \[ 5.5 = (46.0 \sin(35.0))t - \frac{1}{2}(9.81)t^2 \] Let \( v_{y} = 46.0 \sin(35.0) \), then expand to \[ 5.5 = 26.36t - 4.905t^2 \]Rearrange this into standard quadratic form: \[ 4.905t^2 - 26.36t + 5.5 = 0 \]
05

Solve Quadratic Equation

Solve the quadratic equation using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 4.905 \), \( b = -26.36 \), and \( c = 5.5 \). Calculate the discriminant \( b^2 - 4ac \), then solve for \( t \).
06

Evaluate Solutions

Calculate the two possible solutions for \( t \). One will be a negative value which we discard because time cannot be negative. The positive solution is the valid time the ball is in the air.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Initial Velocity
When discussing projectile motion, understanding initial velocity is crucial. It refers to the speed and direction at which an object is launched into the air. In our golf example, the ball is hit with an initial velocity of 46.0 m/s at a 35-degree angle above the horizontal plane.

To solve motion problems, we frequently split the initial velocity into two components:
  • **Horizontal Component (\(v_{x}\)):** This is the part of the velocity that moves along the horizontal plane. It's determined using cosine: \(v_{x} = v \cos \theta\).
  • **Vertical Component (\(v_{y}\)):** This is the part of the velocity directed upwards. It is calculated using sine: \(v_{y} = v \sin \theta\).
For the golf ball:
  • Horizontal velocity: \(v_{x} = 46.0 \times \cos(35.0)\)
  • Vertical velocity: \(v_{y} = 46.0 \times \sin(35.0)\)
These components help us figure out how far and how high the ball will travel.
Kinematic Equation
The kinematic equations describe the motion of objects moving under the influence of forces. They are essential for solving projectile problems.

In this context, we're interested in the equation dealing with vertical motion: \[ y = v_{y} t - \frac{1}{2}gt^2 \]
Here's what each symbol means in our golf ball scenario:
  • \(y\) is the vertical displacement (5.50 m in our case, as the ball travels upward to an elevated green).
  • \(v_{y}\) is the initial vertical velocity, calculated from the initial speed and angle.
  • \(t\) is the time the ball is airborne, the value we're trying to solve.
  • \(g\) is the acceleration due to gravity, acting downward at 9.81 \text{ m/s}^2.
To find \(t\), we substitute the known values into the kinematic equation, ensuring the equation matches the situation.
Quadratic Equation
Solving the kinematic equation led us to a quadratic equation, a key mathematical tool for finding unknown variables in projectile motion.

Our standard quadratic form is:\[ at^2 + bt + c = 0 \]In our exercise, it rearranges to:\[ 4.905t^2 - 26.36t + 5.5 = 0 \]
To solve this, we employ the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • Here, \(a = 4.905\), \(b = -26.36\), and \(c = 5.5\).
  • The discriminant \(b^2 - 4ac\) determines how many solutions exist. A positive discriminant yields two real solutions.
  • We only accept the positive solution because time can't be negative.
By calculating these, we determine how long the golf ball remains in the air before reaching the green.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A ball is thrown upward at a speed \(v_{0}\) at an angle of \(52^{\circ}\) above the horizontal. It reaches a maximum height of 7.5 m. How high would this ball go if it were thrown straight upward at speed v0?

A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40.0 above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

MMH A bird watcher meanders through the woods, walking 0.50 km due east, 0.75 km due south, and 2.15 km in a direction 35.0 north of west. The time required for this trip is 2.50 h. Determine the magnitude and direction (relative to due west) of the bird watcher鈥檚 (a) displacement and (b) average velocity. Use kilometers and hours for distance and time, respectively.

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

As a tennis ball is struck, it departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket?
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Work Energy and Power

Read Explanation

Turning Points in Physics

Read Explanation

Astrophysics

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.