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A golfer hits a shot to a green that is elevated \(3.0 \mathrm{m}\) above the point where the ball is struck. The ball leaves the club at a speed of \(14.0 \mathrm{m} / \mathrm{s}\) at an angle of \(40.0^{\circ}\) 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.

Short Answer

Expert verified
The speed of the ball just before it lands is approximately 14.76 m/s.

Step by step solution

01

Understand the Problem

We need to find the speed of a golf ball just before it lands on a green that is 3.0 meters higher than the point from which it was struck. The ball is launched with an initial speed of 14.0 m/s at an angle of 40.0° to the horizontal. We will use the principles of projectile motion with no air resistance.
02

Break Down the Initial Velocity

Decompose the initial velocity of 14.0 m/s into horizontal and vertical components.The horizontal component is: \\( v_{0x} = 14.0 \, \mathrm{m/s} \times \cos(40.0^\circ) \approx 10.73 \, \mathrm{m/s} \)The vertical component is: \\( v_{0y} = 14.0 \, \mathrm{m/s} \times \sin(40.0^\circ) \approx 8.99 \, \mathrm{m/s} \)
03

Apply Conservation of Energy

Use the principle of conservation of energy that the sum of kinetic and potential energy at launch and at the landing are the same. Calculating only for vertical aspects when using potential energy.Total energy at launch: \\( E_0 = \frac{1}{2} m v_{0y}^2 + mgh_0 \)At launch, height \( h_0 = 0 \): \\( E_0 = \frac{1}{2} m (8.99)^2 \)At landing, the total energy including 3m height increase: \\( E_f = \frac{1}{2} m v_{fy}^2 + mg(3.0) \)Equating both energies: \\( \frac{1}{2} m (8.99)^2 = \frac{1}{2} m v_{fy}^2 + mg(3.0) \)
04

Solve for Final Vertical Speed

Cancel out mass (\(m\)) and solve for \(v_{fy}\):\( \frac{1}{2} (8.99)^2 = \frac{1}{2} v_{fy}^2 + 9.81 \times 3.0 \)\( 80.831 = \frac{1}{2} v_{fy}^2 + 29.43 \) \( 51.401 = \frac{1}{2} v_{fy}^2 \)\( v_{fy}^2 = 102.802 \)\( v_{fy} = \sqrt{102.802} \approx 10.14 \, \mathrm{m/s} \)
05

Determine Total Final Speed

The horizontal speed is constant because there's no air resistance: \( v_{fx} = v_{0x} = 10.73 \, \mathrm{m/s} \).Combine horizontal and vertical final speeds: \\( v_f = \sqrt{v_{fx}^2 + v_{fy}^2} = \sqrt{(10.73)^2 + (10.14)^2} \)Calculate: \\( v_f = \sqrt{115.0529 + 102.8196} = \sqrt{217.8725} \)\( v_f \approx 14.76 \, \mathrm{m/s} \)

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept that's widely used in physics. It states that energy cannot be created or destroyed, only transformed from one form to another. In the context of projectile motion, such as a golf ball being struck, the total energy in the system, which consists of kinetic and potential energy, remains constant if we ignore external forces like air resistance. This means that the energy at the start (when the ball is at its highest point) is equal to the energy just before it hits the ground.

For this projectile: - **Initial energy** comprises the kinetic energy due to the initial velocity and potential energy due to its initial height. - **Final energy** includes the kinetic energy of the motion and potential energy again, but now accounting for the increased elevation of landing 3 meters higher.

By setting these two energies equal, we can solve equations to find unknown values like the speed of the ball just before impact. This is crucial for analyzing projectile motion without getting bogged down by complicated dynamics.
Initial Velocity Components
To analyze projectile motion correctly, it's critical to break down the initial velocity into horizontal and vertical components. This decomposition is based on trigonometry, using the angle and speed at which the projectile is launched.

- The **horizontal component** This is calculated using cosine because it represents movement along the horizontal axis. It's calculated as \( v_{0x} = v_0 \cos(\theta) \), where \( v_0 \) is the initial speed, and \( \theta \) is the angle.- The **vertical component** Similarly, the vertical component is derived from the sine of the angle: \( v_{0y} = v_0 \sin(\theta) \).
For instance, in this problem with an angle of 40 degrees and initial speed of 14 m/s:
  • Horizontal component: \( v_{0x} \approx 10.73 \, \text{m/s} \)
  • Vertical component: \( v_{0y} \approx 8.99 \, \text{m/s} \)
These components are used in subsequent calculations to predict project behavior, such as ranges and time of flight.
Final Speed Calculation
Calculating the final speed of a projectile just before it lands involves determining both its horizontal and vertical speed components. Since horizontal motion is unaffected by gravity, the horizontal speed remains constant. However, for the vertical speed, we account for changes due to potential energy differences and gravity.

In our golf ball problem:
  • **Horizontal Final Speed**: This is the same as its initial horizontal component (\(10.73 \, \text{m/s}\)).
  • **Vertical Final Speed**: From conservation of energy, after accounting for the 3-meter elevation, we find \( v_{fy} \approx 10.14 \, \text{m/s} \).
To find the **total final speed** just before impact, combine these components using the Pythagorean theorem:\[ v_f = \sqrt{v_{fx}^2 + v_{fy}^2} \] Substituting the values: \( v_f = \sqrt{(10.73)^2 + (10.14)^2} \approx 14.76 \, \text{m/s} \)

This value represents the speed with which the ball lands on the elevated green, considering in-flight energy transformations.
Vertical and Horizontal Motion
Projectile motion is a blend of two independent movements happening at the same time: vertical and horizontal. Understanding this separation helps simplify complex paths into more manageable analyses.

- **Vertical Motion**: This part is affected by gravity. For our golfer, the ball starts with a vertical velocity which decreases as it ascends due to gravity until reaching a peak, then increases in speed as it descends. The elevation change demands the inclusion of potential energy differences in calculations. - **Horizontal Motion**: Here, movement occurs at a constant speed as there's no acceleration (ignoring air resistance). This results in a uniform motion parallel to the Earth's surface. By using equations from mechanics, we independently assess the horizontal and vertical components. These motions are the foundation for determining total travel time, range, peak height, and final landing conditions of projectiles—crucial details in understanding dynamic events like the flight of a golf ball.

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Most popular questions from this chapter

The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you lob the ball with an initial speed of \(15.0 \mathrm{m} / \mathrm{s},\) at an angle of \(50.0^{\circ}\) above the horizontal. At this instant your opponent is \(10.0 \mathrm{m}\) away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is \(2.10 \mathrm{m}\) above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.

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.

An eagle is flying horizontally at \(6.0 \mathrm{m} / \mathrm{s}\) with a fish in its claws. It accidentally drops the fish. (a) How much time passes before the fish's speed doubles? (b) How much additional time would be required for the fish's speed to double again?

A spacecraft is traveling with a velocity of \(v_{0 x}=5480 \mathrm{m} / \mathrm{s}\) along the \(+x\) direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the \(+x\) direction of \(a_{x}=1.20 \mathrm{m} / \mathrm{s}^{2}\), while the other gives it an acceleration in the \(+y\) direction of \(a_{y}=8.40 \mathrm{m} / \mathrm{s}^{2}\) At the end of the firing, find (a) \(v_{x}\) and (b) \(v_{y^{*}}\)

A baseball player hits a home run, and the ball lands in the left-field seats, \(7.5 \mathrm{m}\) above the point at which it was hit. It lands with a velocity of \(36 \mathrm{m} / \mathrm{s}\) at an angle of \(28^{\circ}\) below the horizontal (see the drawing). The positive directions are upward and to the right in the drawing. Ignoring air resistance, find the magnitude and direction of the initial velocity with which the ball leaves the bat.

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