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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.

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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.

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