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Chapter 3: Problem 73

A golfer, standing on a fairway, hits a shot to a green that is elevated \(5.50 \mathrm{m}\) above the point where she is standing. If the ball leaves her club with a velocity of \(46.0 \mathrm{m} / \mathrm{s}\) at an angle of \(35.0^{\circ}\) 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 5.17 seconds.

Step by step solution

01

Identify Known Values

We know the initial velocity of the ball \( v_0 = 46.0 \, \mathrm{m/s} \),the angle of projection \( \theta = 35.0^{\circ} \),and the vertical height difference \( \Delta y = 5.50 \, \mathrm{m} \). Gravity \( g \) can be approximated as \( 9.81 \, \mathrm{m/s^2} \).
02

Break Down Initial Velocity

Next, break down the initial velocity into horizontal and vertical components using:\[v_{0x} = v_0 \cdot \cos(\theta)\]\[v_{0y} = v_0 \cdot \sin(\theta)\]For our values:\[v_{0x} = 46.0 \, \mathrm{m/s} \cdot \cos(35.0^{\circ}) = 37.7 \, \mathrm{m/s} \]\[v_{0y} = 46.0 \, \mathrm{m/s} \cdot \sin(35.0^{\circ}) = 26.4 \, \mathrm{m/s} \]
03

Use the Vertical Motion Equation

The vertical motion equation we need is:\[y = v_{0y} \cdot t + \frac{1}{2} a t^2\]Where: \( y = 5.50 \, \mathrm{m} \), \( v_{0y} = 26.4 \, \mathrm{m/s} \),a = \( -9.81 \, \mathrm{m/s^2} \) (entering as negative due to upward motion against gravity).
04

Substitute and Solve for Time

Substitute our known values into the vertical motion equation:\[5.50 = 26.4 \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2\]Reordering gives us:\[4.905 t^2 - 26.4 t + 5.50 = 0\]This is a quadratic equation where:a = 4.905, b = -26.4, c = 5.50.Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we can find the time \( t \).
05

Calculate the Time in the Air

Using the quadratic formula:\[t = \frac{-(-26.4) \pm \sqrt{(-26.4)^2 - 4 \cdot 4.905 \cdot 5.50}}{2 \cdot 4.905}\]\[t = \frac{26.4 \pm \sqrt{696.96 - 107.91}}{9.81}\]\[t = \frac{26.4 \pm \sqrt{589.05}}{9.81}\]\[t = \frac{26.4 \pm 24.27}{9.81}\]This results in two potential solutions:\( t = \frac{50.67}{9.81} \approx 5.17 \, \mathrm{s} \) and \( t = \frac{2.13}{9.81} \approx 0.22 \, \mathrm{s} \).The physical context implies the longer time, 5.17 s, is the correct answer.

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Key Concepts

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

Initial Velocity Components
When dealing with projectile motion, like the golf shot described, understanding initial velocity components is crucial. The initial velocity of the golf ball is given as 46.0 m/s at an angle of 35.0掳 above the horizontal. To make sense of how the ball moves, we need to break this velocity into horizontal and vertical components.
  • The horizontal component is found using: \(v_{0x} = v_0 \cdot \cos(\theta)\), where \(\theta\) is the angle with respect to the ground. For this problem, it calculates to 37.7 m/s. The horizontal motion is consistent because there's no horizontal acceleration assuming negligible air resistance.
  • The vertical component is found using: \(v_{0y} = v_0 \cdot \sin(\theta)\). Here, it calculates to 26.4 m/s. This component is influenced by gravity, affecting the time the ball spends in the air.
Knowing these components helps in understanding how the projectile behaves while moving towards the target.
Quadratic Equation
Understanding quadratic equations is key to solving projectile motion problems such as the one with the golfer. In our scenario, the heights and times are determined by using the quadratic formula.
A quadratic equation typically looks like this: \[ ax^2 + bx + c = 0 \]
For the projectile motion problem, it translates to the form where:
  • \( a = 4.905 \), due to half of gravitational acceleration.
  • \( b = -26.4 \), representing the initial vertical velocity component multiplied by time.
  • \( c = 5.50 \), the vertical displacement.
The solution formula, known as the quadratic formula, is: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In applying this, two time values are found. For the flight of the golf ball, the longer time, 5.17 s, is used because it corresponds with how long the ball is actually in the air before it hits the elevated green.
Vertical Motion Equation
The vertical motion equation plays a significant role in understanding the projectile's motion during the golf shot. This equation is essential because it allows you to determine how long the ball takes to reach its destination at a certain height.
The vertical motion equation is formulated as:\[ y = v_{0y} \cdot t + \frac{1}{2} a t^2 \]Here:
  • \( y = 5.50 \) m, representing the vertical height the ball must cover.
  • \( v_{0y} = 26.4 \) m/s, which is the initial vertical velocity.
  • \( a = -9.81 \) m/s虏, which accounts for the acceleration due to gravity (it's negative because gravity opposes the upward motion).
By substituting these values, the vertical displacement equation becomes a quadratic equation. Solving this helps us find how long the ball stays airborne. This time is essential in determining where and when the ball will land.

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Most popular questions from this chapter

A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the spider is \(0.870 \mathrm{m} / \mathrm{s}\) at an angle of \(35.0^{\circ}\) above the table, and it lands on the magazine \(0.0770 \mathrm{s}\) after leaving the table. Ignore air resistance. How thick is the magazine? Express your answer in millimeters.

A projectile is launched from ground level at an angle of \(12.0^{\circ}\) above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?

On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft has \(x\) and \(y\) components of \(v_{0 x}=4370 \mathrm{m} / \mathrm{s}\) and \(v_{0 y}=6280 \mathrm{m} / \mathrm{s} .\) While the engines are firing, the craft undergoes a displacement that has components of \(x=4.11 \times 10^{6} \mathrm{m}\) and \(y=6.07 \times 10^{6} \mathrm{m} .\) Find the \(x\) and \(y\) components of the craft's acceleration.

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.050 s, during which time it experiences an acceleration of \(340 \mathrm{m} / \mathrm{s}^{2} .\) The ball is launched at an angle of \(51^{\circ}\) above the ground. Determine the horizontal and vertical components of the launch velocity.

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt on which you can either stand or walk. Suppose a speed ramp has a length of \(105 \mathrm{m}\) and is moving at a speed of \(2.0 \mathrm{m} / \mathrm{s}\) relative to the ground. In addition, suppose you can cover this distance in 75 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the \(105 \mathrm{m}\) using the speed ramp?
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