Based on the measured angles of the marble launcher, calculate the initial horizontal and vertical components of the marble’s velocity for each angle, assuming the velocity vector has the same magnitude as you calculated from Part I (4.31 m/s). angle 1: θ = 15° -->initial height 0.027m, horizontal distance 0.145m angle 2: θ = 45° -->initial height 0.027m, horizontal distance 0.209m angle 3: θ = 70° --> initial height 0.027m, horizontal distance 0.118m
Based on the measured angles of the marble launcher, calculate the initial horizontal and vertical components of the marble’s velocity for each angle, assuming the velocity vector has the same magnitude as you calculated from Part I (4.31 m/s). angle 1: θ = 15° -->initial height 0.027m, horizontal distance 0.145m angle 2: θ = 45° -->initial height 0.027m, horizontal distance 0.209m angle 3: θ = 70° --> initial height 0.027m, horizontal distance 0.118m
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter1: Introduction And Vectors
Section: Chapter Questions
Problem 50P: Consider the three displacement vectors A=(3i3j)m, B=(i4j)m, and C=(2i+5j)m. Use the component...
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Based on the measured angles of the marble launcher, calculate the initial horizontal and vertical components of the marble’s velocity for each angle, assuming the velocity vector has the same magnitude as you calculated from Part I (4.31 m/s).
angle 1: θ = 15° -->initial height 0.027m, horizontal distance 0.145m
angle 2: θ = 45° -->initial height 0.027m, horizontal distance 0.209m
angle 3: θ = 70° --> initial height 0.027m, horizontal distance 0.118m
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