A projectile is launched with an initial velocity 38 m/s and an angle from the horizontal of 45°. The acceleration due to gravity is 9.81 m/s². Assume the projectile is launched from the surface of the Earth, i.e.. at y (0) = 0. then the altitude of the projectile as a function of time is: y(t) = gt² + vo sin(0)t Where vo is the initial velocity. What is the maximum altitude achieved by the projectile?

A projectile is launched with an initial velocity 38 m/s and an angle from the horizontal of 45°. The acceleration due to gravity is 9.81 m/s². Assume the projectile is launched from the surface of the Earth, i.e.. at y (0) = 0. then the altitude of the projectile as a function of time is: y(t) = gt² + vo sin(0)t Where vo is the initial velocity. What is the maximum altitude achieved by the projectile?

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Description: A projectile is launched with an initial velocity 38 m/s and an angle from the horizontalof 45°. The acceleration due to gravity is 9.81 m/s². Assume the projectile is launchedfrom the surface of the Earth. i.e., at y (0) = 0. then the altitude of t

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter2: Vectors

Section: Chapter Questions

Problem 5CQ: Suppose you add two vectors A and B . What relative direction between them produces the resultant...

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