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ghtward at an angle of 17°. What is the high jumper's initial h elocity, in m/s? nswer: m/s

ghtward at an angle of 17°. What is the high jumper's initial h elocity, in m/s? nswer: m/s

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement:** A high jumper leaves the ground with an initial velocity of 11.1 m/s rightward at an angle of 17°. What is the high jumper’s initial horizontal velocity, in m/s? **Answer:** _______ m/s **Explanation:** To find the initial horizontal velocity, use the formula for the horizontal component of velocity: \[ v_{\text{horizontal}} = v \cdot \cos(\theta) \] Where: - \( v \) is the initial velocity (11.1 m/s), - \( \theta \) is the angle with respect to the horizontal (17°), - \( \cos \) is the cosine function. By calculating the cosine of 17° and multiplying it by 11.1 m/s, you can find the initial horizontal velocity.
Transcribed Image Text:**Problem Statement:** A high jumper leaves the ground with an initial velocity of 11.1 m/s rightward at an angle of 17°. What is the high jumper’s initial horizontal velocity, in m/s? **Answer:** _______ m/s **Explanation:** To find the initial horizontal velocity, use the formula for the horizontal component of velocity: \[ v_{\text{horizontal}} = v \cdot \cos(\theta) \] Where: - \( v \) is the initial velocity (11.1 m/s), - \( \theta \) is the angle with respect to the horizontal (17°), - \( \cos \) is the cosine function. By calculating the cosine of 17° and multiplying it by 11.1 m/s, you can find the initial horizontal velocity.
Expert Solution
Step 1

Given

Initial velocity , u = 11.1 m/s

Angle, θ  = 17o

 

 

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