SOLUTION: An object is launched from ground level directly upward at a rate of 144 feet per second. The equation for the object's height is h(x)=-16x^2+144x. What is the value of x, is the

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Question 1130551: An object is launched from ground level directly upward at a rate of 144 feet per second. The equation for the object's height is h(x)=-16x^2+144x. What is the value of x, is the object at or above a height of 288 feet? How long is the object at or above this height?
Answer by ikleyn(52946)   (Show Source): You can put this solution on YOUR website!
.
-16x^2 + 144x = 288      (1)     ====>


16x^2 - 144x + 288 = 0  is equivalent to


x^2 - 9x + 18 = 0


(x-3)*(x-6) = 0.


The roots of the equation (1)  are  3  and  6  seconds.


So, the object is above the height of 144 ft  from  the time moment of 3 seconds till the time moment of 6 seconds,

or exactly  3 = 6 - 3 seconds.           ANSWER

Solved.

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In this site, there is a bunch of lessons on a projectile thrown/shot/launched vertically up
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


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