### Physics Problem: Skier's Jump#### Problem Statement:The skier starts from rest at point \( A \). Determine the horizontal distance \( X \) traveled after he leaves the ramp and the landing speed. Neglect friction.#### Diagram Explanation:The provided diagram shows a side view schematic of a skier's jump. Here's a step-by-step explanation of the diagram:1. **Starting Point (A)**: - The skier starts at a height of 25 meters above the ground.2. **Inclined Ramp (Angle of 20°)**: - The skier begins at point \( A \), moves down an inclined ramp making an angle of 20° with the horizontal.3. **Ramp End Point (B)**: - The end of the ramp is labeled as point \( B \).4. **Parabolic Trajectory (x, y Coordinates)**: - After leaving the ramp, the skier follows a parabolic trajectory. - The equation for the parabola is \( y = -5 - 0.01x^2 \), where \( y \) is the vertical distance and \( x \) is the horizontal distance.5. **Distance to Landing Point (C)**: - The horizontal distance \( X \) between the ramp end point \( B \) and the landing point \( C \) is to be determined.#### Given Options:You need to choose one correct pair for the horizontal distance \( X \) and the landing speed \( v_c \) from the options below:1. \( X = 8.96 \, \text{m}, \quad v_c = 21.2 \, \frac{\text{m}}{\text{s}} \)2. \( X = 13.1 \, \text{m}, \quad v_c = 24.9 \, \frac{\text{m}}{\text{s}} \)3. \( X = 10.2 \, \text{m}, \quad v_c = 24.2 \, \frac{\text{m}}{\text{s}} \)4. \( X = 7.81 \, \text{m}, \quad v_c = 18.1 \, \frac{\text{m}}{\text{s}} \)#### Instructions:Select the correct option for the horizontal distance and the landing speed based on the given data. Click "Submit" once

Answered: Image /qna-images/answer/4cd47cd6-0d53-4fcb-b589-a49bbcb05b71.jpg

Part A The Skier starts from rest at point A. Determine the horizontal distance X traveled after he leaves the ramp and the landing speed. Neglect friction h O X 8.96 m, vc = -21.2 OX= 13.1 m, vc = 24.9 m S OX= 10.2 m, vc = 24.2 O X= 7.81 m, ve = 18.1 Submit mn S Request Answer m S S 1 I A S 25 m 20° B 5 m X C y=-5-0.01² ENG

Part A The Skier starts from rest at point A. Determine the horizontal distance X traveled after he leaves the ramp and the landing speed. Neglect friction h O X 8.96 m, vc = -21.2 OX= 13.1 m, vc = 24.9 m S OX= 10.2 m, vc = 24.2 O X= 7.81 m, ve = 18.1 Submit mn S Request Answer m S S 1 I A S 25 m 20° B 5 m X C y=-5-0.01² ENG

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: The thrust on the 5-Mg rocket sled is shown in the graph. Neglect friction. (Figure 1) T (KN) 20-…

Q: The boats A and B travel with constant speeds of v₁ = 16 m/s and "B" 11 m/s when they leave the pier…

Q: Blocks A and B each weigh 0.8 lbs. And Block C weighs 2.4 lbs. The coefficient of static friction…

A: applying principle of work and energy to block A at both the position 1 & 2 before the impact…

Q: This is a multi-part question. Once an answer is submitted, you will be unable to return the rope is…

A: The angle of rope is .The weight of sphere A is .The weight of sphere B is .The speed of sphere A is…

Q: A rocket runs out of fuel in the position shown and continues in unpowered flight above the…

Q: Q5. As shown in the image below, the 54-kg skier slides down the smooth slope for which the shape is…

A: Given data: m=54 kgy=0.03x2+6 mx=10 mv=7 m/s Need to determine the normal force exerted on at point…

Q: An engineer wants to design an oval racetrack such that 3.20 × 10³ lb racecars can round the exactly…

A: To Find : The banking angle θ necessary for the race cars to navigate the turns at 103 mi/h without…

Q: Current Attempt in Progress The system starts from rest with no slack in the cable. The block has a…

Q: Requirea information NOTE: This is a multi-part question. Once an answer is submitted, you will be…

A: The mass of the sphere A is The initial speed of sphere A is The coefficient of restitution is The…

Q: cc BY UBC Engineering The 3 kg block sits on the edge of a 1 m disk that is accelerating with a rope…

A: Given Mass, m = 3 kg Diameter, d = 1 m Velocity, v = 7 m/s Find The tension,…

Q: Part A. Determine the minimum height h of the hill crest so that the car travels around both inside…

A: Find the hight at point A and normal reaction on the car at point B and point C.

Q: During that move, how hard was Sofia pulling Boris?

A: we are to determine the reaction on boris in this problem.

Q: The free-rolling ramp has a mass of 40 kg. A 10-kg crate is released from rest at A and slides down…

A: The mass of the crate is m=10 kg The mass of ramp is M=40 kg The system is at rest at point A.

Q: Part A Determine the time needed for the load at B to attain a speed of 8 m/s, starting from rest…

Q: A 5kg block is pushed up along the slope as shown. The coefficient of kinetic friction is uk =0.2.…

Q: Compute the acceleration of block A for the instant depicted. Neglect the masses of the pulleys. 39…

Q: Please show complete work in handwritten format, Do not use Chatgpt.

A: Step 1:

Q: 6) An object falls a distance h from rest. If it travels 0.50h in the last 1.00 s, find (a) the time…

A: Given Datas=0.5hg=9.81 m/s2

Q: The Skier starts from rest at point A. Determine the horizontal distance X traveled after he leaves…

Concept explainers

Applied Fluid Mechanics

A fluid is a substance that is continuously deformed by the application of shear stress. The rate of deformation of a fluid depends on its viscosity. The fluid tries to flow when it interacts with some other system.

Question

### Physics Problem: Skier's Jump

#### Problem Statement:
The skier starts from rest at point \( A \). Determine the horizontal distance \( X \) traveled after he leaves the ramp and the landing speed. Neglect friction.

#### Diagram Explanation:
The provided diagram shows a side view schematic of a skier's jump. Here's a step-by-step explanation of the diagram:

1. **Starting Point (A)**:
- The skier starts at a height of 25 meters above the ground.

2. **Inclined Ramp (Angle of 20°)**:
- The skier begins at point \( A \), moves down an inclined ramp making an angle of 20° with the horizontal.

3. **Ramp End Point (B)**:
- The end of the ramp is labeled as point \( B \).

4. **Parabolic Trajectory (x, y Coordinates)**:
- After leaving the ramp, the skier follows a parabolic trajectory.
- The equation for the parabola is \( y = -5 - 0.01x^2 \), where \( y \) is the vertical distance and \( x \) is the horizontal distance.

5. **Distance to Landing Point (C)**:
- The horizontal distance \( X \) between the ramp end point \( B \) and the landing point \( C \) is to be determined.

#### Given Options:
You need to choose one correct pair for the horizontal distance \( X \) and the landing speed \( v_c \) from the options below:

1. \( X = 8.96 \, \text{m}, \quad v_c = 21.2 \, \frac{\text{m}}{\text{s}} \)
2. \( X = 13.1 \, \text{m}, \quad v_c = 24.9 \, \frac{\text{m}}{\text{s}} \)
3. \( X = 10.2 \, \text{m}, \quad v_c = 24.2 \, \frac{\text{m}}{\text{s}} \)
4. \( X = 7.81 \, \text{m}, \quad v_c = 18.1 \, \frac{\text{m}}{\text{s}} \)

#### Instructions:
Select the correct option for the horizontal distance and the landing speed based on the given data. Click "Submit" once

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

1 Problem During a particularly snowy winter in Charlotte, a child of mass m slides down a ramp starting at rest from point A and encounters the melted grassy area at B before coming to a full stop at C. Assume the ice is frictionless. The height of point A is hĄ and the distance between B and C is L. What is the coefficient of friction of the grass, k? Assume that g = 9.81 m/s², ha = 5 m, m = 20 kg, and L = 10 m. hA P JA Ice B Grass L C
Q23. As shown in the image below, the force acting on the 5-kg crate is a function of time. The coefficient of kinetic friction between the crate and the surface is uk= 0.14. Determine the crate's speed at t= 2.1 s if its initial speed v1 = 2.9 m/s. Please pay attention: the numbers may change since they are randomized. Your answer must include 2 places after the decimal point, and proper Sl unit. Take g = 9.81 m/s2. F = (201² +30)N (t in second) F. 30 V1 Your Answer: Answer units
In the following problem please include a sketch of the problem, quantities given, quantities to be found, and boxed answer with units.
Label steps clearly
Q24. The buoyancy force on the 580-kg balloon is F = 6 kN, and the air force acting {C₁.ti-C₂. √t j N, when t is time in on the balloon is Fair seconds, constants C₁ = 24 and C₂ = 240. If the balloon starts from rest, determine its speed after 28 seconds. Neglect the size of the balloon but don't neglect its weight. Please pay attention: the numbers may change since they are randomized. Your answer must include 2 places after the decimal point, and proper SI unit. Take g = 9.81 m/s². Fa air Your Answer: Answer = F = 6 kN units
I need a clear answer by hand, not by keyboard and fast answer within 20 minutes. Thank you | dybala The system is initially at rest with no slack in the cable, and the mass and friction of the pulleys are negligible. Determine the following: 1- The acceleration of the block if (a) T = 20 N and (b) T= = 30 N. 2- The velocity of the block in 4 seconds if T = 30 N. Hs=0.50 H=0.40) 15 kg 30° T