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A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 155 feet per second from an initial height of 59 feet off the ground, then the height of the projectile, h, in feet, t seconds after it's shot is given by the equation: h=-16t² + 155t +59 Find the two points in time when the object is 95 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places). Answer: The object is 95 feet off the ground at the following times: Enter your two answers separated by a comma.

A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 155 feet per second from an initial height of 59 feet off the ground, then the height of the projectile, h, in feet, t seconds after it's shot is given by the equation: h=-16t² + 155t +59 Find the two points in time when the object is 95 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places). Answer: The object is 95 feet off the ground at the following times: Enter your two answers separated by a comma.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 155 feet per second from an initial height of 59 feet off the ground, then the height of the projectile, h, in feet, t seconds after it's shot is given by the equation: h=-16t2 + 155t +59 Find the two points in time when the object is 95 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places). Answer: The object is 95 feet off the ground at the following times: Enter your two answers separated by a comma.
Transcribed Image Text:A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 155 feet per second from an initial height of 59 feet off the ground, then the height of the projectile, h, in feet, t seconds after it's shot is given by the equation: h=-16t2 + 155t +59 Find the two points in time when the object is 95 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places). Answer: The object is 95 feet off the ground at the following times: Enter your two answers separated by a comma.
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