(a) Explanation Given: With an initial velocity of 96 feet per second, a ball is thrown perpendicularly upward from ground level. The acceleration of the ball is a ( t ) = − 32 feet/second 2 . (Air resistanceis neglected.) Formula used: The velocity of the ball for the utmost height at time t ′ must be 0, thus, v ( t ′ ) = 0 Calculation: Suppose that h ( t ) be the height of the ball from the ground level at any time t in sec. Since, the ball is thrown vertically upward from ground level with an initial velocity of 96 feet per secondthen the initial condition at t = 0 is: h ( 0 ) = 0 …… (1) v ( 0 ) = d h d t = 96 …… (2) Since the ball’s acceleration is a ( t ) = − 32 feet/sec 2 , therefore, the equation of the motion is as following: a ( t ) = d v d t = − 32 Performing integration with respect to t , ∫ d v = − 32 ∫ d t + c v ( t ) = − 32 t + c Implementing the initial condition (2), v ( 0 ) = − 32 ( 0 ) + c 96 = c Hence, the velocity of the ball at any time t is: d h d t = − 32 t + 96 … (b) To determine To calculate: After how many seconds the ball velocity will be one-half of the initial velocity when the velocity of the ball at any time t sec is v ( t ) = − 32 t + 96 . (c) To determine To calculate: The ball height when its velocity is one-half the initial velocitywhen the height of the ball at any time t sec is h ( t ) = − 16 t 2 + 96 t and the velocity of the ball is one-half the initial velocity after 1.5 sec.

11th Edition

ISBN: 9781337275361

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 4, Problem 13RE

(a)

To determine

To calculate: The utmost height that the ball canreach and the time the ball takes to attain its utmost height. The given initial velocity is 96 feet per second and the ball’s acceleration is a(t)=−32 feet/second²

(b)

To determine

To calculate: After how many seconds the ball velocity will be one-half of the initial velocity when the velocity of the ball at any time t sec is v(t)=−32t+96.

(c)

To determine

To calculate: The ball height when its velocity is one-half the initial velocitywhen the height of the ball at any time t sec is h(t)=−16t2+96t and the velocity of the ball is one-half the initial velocity after 1.5 sec.

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Chapter 4, Problem 12RE

Chapter 4, Problem 14RE

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The utmost height that the ball canreach and the time the ball takes to attain its utmost height. The given initial velocity is 96 feet per second and the ball’s acceleration is a ( t ) = − 32 feet/second 2

Solution Summary: The author calculates the utmost height that the ball can reach and the time it takes to attain it.

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To calculate: The utmost height that the ball canreach and the time the ball takes to attain its utmost height. The given initial velocity is 96 feet per second and the ball’s acceleration is a(t)=−32 feet/second²

To calculate: After how many seconds the ball velocity will be one-half of the initial velocity when the velocity of the ball at any time t sec is v(t)=−32t+96.

To calculate: The ball height when its velocity is one-half the initial velocitywhen the height of the ball at any time t sec is h(t)=−16t2+96t and the velocity of the ball is one-half the initial velocity after 1.5 sec.

Want to see the full answer?

Chapter 4 Solutions

The utmost height that the ball canreach and the time the ball takes to attain its utmost height. The given initial velocity is 96 feet per second and the ball’s acceleration is a ( t ) = − 32 feet/second 2

Solution Summary: The author calculates the utmost height that the ball can reach and the time it takes to attain it.

Concept explainers

Videos

To calculate: The utmost height that the ball canreach and the time the ball takes to attain its utmost height. The given initial velocity is 96 feet per second and the ball’s acceleration is a(t)=−32 feet/second2

To calculate: After how many seconds the ball velocity will be one-half of the initial velocity when the velocity of the ball at any time t sec is v(t)=−32t+96.

To calculate: The ball height when its velocity is one-half the initial velocitywhen the height of the ball at any time t sec is h(t)=−16t2+96t and the velocity of the ball is one-half the initial velocity after 1.5 sec.

Want to see the full answer?

Chapter 4 Solutions

To calculate: The utmost height that the ball canreach and the time the ball takes to attain its utmost height. The given initial velocity is 96 feet per second and the ball’s acceleration is a(t)=−32 feet/second²