A ball is launched straight up in the air from a height of 5 feet. Its velocity (feet/second) t seconds after launch is given by f(t) = - 32t + 286. Find its height 6 seconds after launch. The height of the ball 6 seconds after launch is feet. (Round answer to nearest tenth.)

A ball is launched straight up in the air from a height of 5 feet. Its velocity (feet/second) t seconds after launch is given by f(t) = - 32t + 286. Find its height 6 seconds after launch. The height of the ball 6 seconds after launch is feet. (Round answer to nearest tenth.)

Elementary Algebra

17th Edition

ISBN:9780998625713

Author:Lynn Marecek, MaryAnne Anthony-Smith

Publisher:Lynn Marecek, MaryAnne Anthony-Smith

Chapter10: Quadratic Equations

Section10.4: Solve Applications Modeled By Quadratic Equations

Problem 159E: A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial...

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A ball is launched straight up in the air from a height of 5 feet. Its velocity (feet/second) t seconds after launch is given by f(t) = - 32t + 286. Find its height 6 seconds
after launch.
The height of the ball 6 seconds after launch is feet.
(Round answer to nearest tenth.)

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