/*! 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 22 The 1994 Winter Olympics include... [FREE SOLUTION] | 91Ó°ÊÓ

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The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers 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. In the women's competition, the end of a typical 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 17.96 m/s.

Step by step solution

01

Identify initial data and target

We need to find the skier's launch speed. The problem gives us the launch angle of \(63^{\circ}\) and the maximum height attained by the skier is \(13 \mathrm{~m}\).
02

Analyze the vertical motion

Use the kinematic equation for vertical motion: \[ v_{y}^2 = v_{0_y}^2 + 2a y \]Here, the final vertical velocity \(v_{y} = 0\) m/s at the peak of the jump, \(a = -9.8 \mathrm{~m/s^2}\) (acceleration due to gravity), and \(y = 13 \mathrm{~m}\).
03

Set up the equation for vertical motion

Substitute the known values into the equation:\[ 0 = (v_{0_y})^2 + 2(-9.8)(13) \]Simplify to find \(v_{0_y}\).
04

Calculate initial vertical speed

Solve for \(v_{0_y}\): \[ (v_{0_y})^2 = 2 \times 9.8 \times 13 \]\[ v_{0_y} = \sqrt{2 \times 9.8 \times 13} \]\[ v_{0_y} \approx 15.99 \mathrm{~m/s} \].
05

Relate launch speed to vertical speed

The vertical component of the launch speed \(v_{0_y}\) is related to the launch speed \(v_0\) by the angle:\[ v_{0_y} = v_0 \sin(63^{\circ}) \]
06

Solve for skier's launch speed

Rearrange the equation to solve for \(v_0\):\[ v_0 = \frac{v_{0_y}}{\sin(63^{\circ})} \]\[ v_0 = \frac{15.99}{\sin(63^{\circ})} \]Calculate \(v_0\) to find the launch speed.

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Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools used to analyze the motion of objects, such as skiers in the air during an Olympic event. These equations allow us to connect various physical quantities like velocity, acceleration, and displacement over time.
The primary kinematic equation used in analyzing vertical motion for projectile motion is: \[ v_{y}^2 = v_{0_y}^2 + 2a y \]
  • \( v_{y} \) represents the final vertical velocity, which is 0 m/s at the peak of motion.
  • \( v_{0_y} \) is the initial vertical component of the velocity.
  • \( a \) represents the acceleration, in this case, due to gravity, which is \(-9.8 \mathrm{~m/s^2}\).
  • \( y \) is the vertical displacement, which, in this problem, is 13 meters.
In the specified scenario, the kinematic equation is used to determine the initial vertical velocity, which is a stepping stone toward finding the overall launch speed. This makes understanding the motion of projectiles in a real-world application much more structured and predictable, allowing us to calculate complex maneuvers like those performed by skiers.
Launch Angle
The launch angle in projectile motion defines the initial direction of an object. In this particular exercise, the launch angle is set at \(63^{\circ}\) above the horizontal. This angle significantly influences the trajectory and the distribution of the initial velocity into horizontal and vertical components.

Understanding the implications of launch angles is crucial because:
  • A larger launch angle will typically result in a higher and shorter trajectory.
  • Conversely, a smaller angle might yield a lower and longer path.
  • The angle impacts how high and how far the skier will travel through the air.
In the step-by-step solution, the angle is critical for calculating the vertical component of the initial velocity \( v_{0_y} \). The vertical motion equation and the launch angle work together to give a clear picture of the skier's motion. This knowledge is pivotal in various applications beyond skiing, such as launching projectiles in physics or engineering contexts.
Vertical Velocity
The concept of vertical velocity is central to solving this projectile motion problem. It refers to the component of velocity that acts along the vertical axis, separate from any horizontal motion.
When discussing projectiles like our skier, it's essential to distinguish between vertical and horizontal velocities due to their unique behaviors influenced by forces like gravity.
Here's why vertical velocity matters:
  • At the maximum height, the vertical velocity \( v_{y} \) becomes 0 m/s as gravity slows the upward motion and eventually reverses it.
  • The initial vertical velocity \( v_{0_y} \) can be calculated using the known maximum height and gravitational acceleration.
  • This value forms a critical piece of information for ultimately determining the launch speed \( v_0 \).
Given the launch angle and the kinematic equations, we can break down the launch speed into its components, tackling the vertical and horizontal independently. This clearer understanding of vertical velocity not just aids in solving problems about projectile motion but also extends into real-world applications, influencing designs and predictions.

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Most popular questions from this chapter

The altitude of a hang glider is increasing at a rate of \(6.80 \mathrm{~m} / \mathrm{s}\). At the same time, the shadow of the glider moves along the ground at a speed of \(15.5 \mathrm{~m} / \mathrm{s}\) when the sun is directly overhead. Find the magnitude of the glider's velocity.

The captain of a plane wishes to proceed due west. The cruising speed of the plane is \(245 \mathrm{~m} / \mathrm{s}\) relative to the air. A weather report indicates that a \(38.0-\mathrm{m} / \mathrm{s}\) wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane relative to the air?

A rocket is fired at a speed of \(75.0 \mathrm{~m} / \mathrm{s}\) from ground level, at an angle of \(60.0^{\circ}\) above the horizontal. The rocket is fired toward an 11.0 -m-high wall, which is located \(27.0 \mathrm{~m}\) away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by \(1.3 \mathrm{~m}\). A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is \(1.0 \mathrm{~m}\) above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is \(1.7 \mathrm{~m}\) above the ground. Finally, he leaps back to the other tree, now landing at a spot that is \(2.5 \mathrm{~m}\) above the ground. What is the magnitude of the squirrel's displacement?

(a) Does Barbara see him moving toward the east or toward the west? (b) Does Barbara see him moving toward the north or toward the south? (c) Considering your answers to parts (a) and (b), how does Barbara see Neil moving relative to herself, toward the east and north, toward the east and south, toward the west and north, or toward the west and south? Justify your answers in each case. With respect to the ground, Barbara is skating due south at a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). With respect to the ground, Neil is skating due west at a speed of \(3.2 \mathrm{~m} / \mathrm{s} .\) Find Neil's velocity (magnitude and direction relative to due west) as seen by Barbara. Make sure that your answer agrees with your answer to part (c) of the Concept Questions.

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