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Chapter 3: Problem 21

In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. The end of a launch ramp is directed \(63^{\circ}\) above the horizontal. With this launch angle, a skier attains a height of \(13 \mathrm{m}\) above the end of the ramp. What is the skier's launch speed?

Short Answer

Expert verified
The skier's launch speed is approximately 17.8 m/s.

Step by step solution

01

Understand the Problem

We need to find the skier's launch speed based on the launch angle and maximum height attained. The given launch angle is \(63^{\circ}\), and the maximum height reached is \(13 \, \text{m}\).
02

Apply the Basic Projectile Motion Formulas

In projectile motion, the vertical motion can be described using the formula: \[v_{Oy}^2 = v_O^2 \sin^2 \theta - 2g h_{max},\]where \(v_{Oy}\) is the initial vertical velocity, \(v_O\) is the launch speed, \(\theta\) is the launch angle, \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)), and \(h_{max}\) is the maximum height.
03

Identify Initial Vertical Velocity

Since the maximum height is reached, the final vertical velocity at the peak is \(0 \, \text{m/s}\). Thus, the vertical component of the initial velocity is given by \[0 = v_O^2 \sin^2 \theta - 2g h_{max}.\]
04

Solve for Launch Speed

Rearrange the formula to solve for \(v_O\): \[v_O^2 \sin^2 \theta = 2g h_{max}.\]Plug in the values: \[v_O^2 \cdot \sin^2 (63^{\circ}) = 2 \cdot 9.81 \, \text{m/s}^2 \cdot 13 \, \text{m}.\]Calculate \(v_O\) by evaluating this expression:\[v_O = \sqrt{\frac{2 \cdot 9.81 \, \text{m/s}^2 \cdot 13 \, \text{m}}{\sin^2 (63^{\circ})}}.\]
05

Calculate Sin of the Angle

Calculate \(\sin(63^{\circ})\), which is approximately \(0.8910\).
06

Final Computation

Substitute \(\sin(63^{\circ})\) back into the equation:\[v_O = \sqrt{\frac{2 \cdot 9.81 \, \text{m/s}^2 \cdot 13}{0.8910^2}}.\]Compute this to find \(v_O\).
07

Result Interpretation

After calculating, you find \(v_O \approx 17.8 \, \text{m/s}\). This is the launch speed of the skier.

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Key Concepts

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

Launch Speed
Launch speed is a crucial concept in projectile motion. It is the speed at which the object begins its journey, propelling from its initial position. In the context of aerial skiing, this refers to how fast the skier leaves the ramp. Understanding launch speed involves:
  • Recognizing it as the magnitude of the initial velocity vector (\( v_O \)), a combination of both horizontal and vertical components.
  • It dictates how far and high a skier can travel, impacting their overall performance in the air.
To determine the launch speed, we can use essential physics formulas. In this exercise, we rearrange the projectile motion equation that accounts for the vertical component. By considering the maximum height and entry angles, we derive the skier's initial speed as approximately 17.8 m/s. This calculation shows the necessity of precise speed to achieve the desired height in the aerial performance.
Maximum Height
The maximum height in projectile motion refers to the peak altitude an object reaches in its trajectory. For a skier, this would be the highest point they reach above the ramp. Calculating maximum height involves several factors:
  • The angle of launch, which in this problem is \(63^{\circ}\).
  • The gravitational pull, generally considered as \(9.81 \text{ m/s}^2\), dragging the skier back down.
In our exercise, the problem states the skier reaches an altitude of 13 meters. From a physics standpoint, at this peak, the skier's vertical velocity momentarily becomes zero before gravity pulls them back downward. Understanding maximum height helps determine how high maneuvers can be executed in the air, directly influencing the complexity and type of acrobatics performed.
Initial Vertical Velocity
Initial vertical velocity is a vital component in comprehending projectile motion. It represents the initial velocity component that acts in the vertical direction when an object is launched. For the skiing athlete, it is how fast they are moving upward off the ramp at the moment of takeoff.Breaking down initial vertical velocity:
  • It is derived from the overall launch speed and the sine of the launch angle (\( \theta \)). The formula is:\[ v_{Oy} = v_O \sin(\theta). \]
  • This value is crucial to calculate since it dictates how quickly the object will start its vertical climb toward the maximum height.
While at the maximum height, the initial vertical velocity helps analyze how high the skier can ascend before succumbing to gravity. It is an essential determinant in sports involving high-flying dynamics like aerial skiing, affecting trajectory and performance.

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Most popular questions from this chapter

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.

Multiple-Concept Example 4 provides useful background for this problem. A diver runs horizontally with a speed of \(1.20 \mathrm{m} / \mathrm{s}\) off a platform that is \(10.0 \mathrm{m}\) above the water. What is his speed just before striking the water?

A space vehicle is coasting at a constant velocity of \(21.0 \mathrm{m} / \mathrm{s}\) in the \(+y\) direction relative to a space station. The pilot of the vehicle fires a \(\mathrm{RCS}\) (reaction control system) thruster, which causes it to accelerate at \(0.320 \mathrm{m} / \mathrm{s}^{2}\) in the \(+x\) direction. After \(45.0 \mathrm{s},\) the pilot shuts off the \(\mathrm{RCS}\) thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle measured from the \(+y\) direction.

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.)

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