Explanation Given: If an object is shot upward from the ground with an initial velocity of 256 ft/sec, its distance in feet above the ground at the end of t seconds is given by s ( t ) = 256 t − 16 t 2 . At highest point the velocity of shot becomes 0. The given equation is distance in term of t The derivative of distance, s ( t ) , gives velocity of object. Therefore, v ( t ) = s ' ( t ) v ( t ) = 256 − 32 t At highest point velocity is 0 So, 256 − 32 t = 0

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 1, Problem 56SP

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To find: The highest point for s(t)=256t−16t2 projectile.

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Chapter 1, Problem 55SP

Chapter 1, Problem 57SP

Chapter 1 Solutions

Calculus

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The highest point for s ( t ) = 256 t − 16 t 2 projectile.

Solution Summary: The author explains that the highest point for s(t)=256t-16t 2 projectile is 1024 ft.

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To find: The highest point for s(t)=256t−16t2 projectile.

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Chapter 1 Solutions