Question 1183499: In a recent eruption, a projectile is ejected within an initial velocity of 400 ft per second. Where t is time in seconds, use the equation (-16t^2 + 208t) and steps taken to determine the following:
1. the time that it takes the projectile to reach its highest point,
2. the maximum height of the projectile,
3. the time it takes the projectile to return to the ground.
Answer by ikleyn(52795)   (Show Source):
You can put this solution on YOUR website! .
In a recent eruption, a projectile is ejected within an initial velocity of 400 ft per second.
Where t is time in seconds, use the equation (-16t^2 + 208t) and steps taken to determine the following:
1. the time that it takes the projectile to reach its highest point,
2. the maximum height of the projectile,
3. the time it takes the projectile to return to the ground.
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In this post, the written "height" function DOES NOT CORRESPOND to the described problem.
The term " + 208t " is not consisted with the initial velocity of 400 ft/s.
From your post, I conclude that you are UNFAMILIAR with the forms of the projectile equation.
To learn this subject, read this introductory info below.
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Introduction written specially for beginners who don't know the subject AT ALL
The formula in your post is written incorrectly. FATALLY INCORRECTLY, it is what I want to say.
I observed it many times in this forum that the students / (the visitors) write this formula incorrectly.
because they do not understand the meaning of its terms.
So I prepared this text below as a standard introduction to the subject for such students / visitors.
Be calm, you are in good hands, and read my post to the very end.
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If you have the formula for a height given to you as a function of time in the form
h(t) = -at^2 + bt + c, (1)
where "a", "b" and "c" are real numbers, a > 0, then in this formula
(a) the initial height is equal to the coefficient "c" value;
(b) the initial velocity is the coefficient "b" in the formula;
(c) the coefficient "a" value is half of the gravity acceleration.
For the Earth conditions, the gravity acceleration is g = 32 ft/s^2,
if you use feet for height.
So, in this case a =  = 32/2 = 16 ft/s^2 (the numerical value).
(d) To find the height at the time moment "t", simply substitute the value of "t" into the formula (1) and calculate.
(e) To find the time "t" when the height has a given value h =  , substitute h = into equation (1)
and solve equation
h(t) = -at^2 + bt + c = . (2)
(f) To find the time when the height is maximal, use the formula
 =  . (3)
(g) To find the maximal height, substitute the time value t= of the formula (3) into the formula (1).
What's all you need to know.
To answer your questions, use my instructions and make calculations on your own.
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To see numerous examples of solved problems, look into the lessons
- 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
in this site.
Consider these lessons as your textbook, handbook, tutorials and (free of charge) home teacher.
Read them attentively and learn how to solve this type of problems once and for all.
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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If this introduction is helpful to you, I will be happy.
If it will be not enough to you to solve the problem, come again,
but with one indispensable condition: your equation MUST be written correctly.
Don't forget to post your "THANKS" to me for my teaching (!)
Come again, if you need.
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