Question 1192009: Got half of a problem done not sure how to complete it,
A baseball is hit so that its height in feet after t seconds is s(t)=-4t^2+38t+3 .
(a) How high is the baseball after 1 second?
37ft
(b) Find the maximum height of the baseball.
?
Any explanation is very appreciated! thank you so much
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52797)   (Show Source):
You can put this solution on YOUR website! .
Got half of a problem done not sure how to complete it,
A baseball is hit so that its height in feet after t seconds is s(t)=-4t^2+38t+3 .
(a) How high is the baseball after 1 second?
37ft
(b) Find the maximum height of the baseball.
?
Any explanation is very appreciated! thank you so much
~~~~~~~~~~~~~~~
As it is written, this formula does not work.
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.
---------------
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 = 9.81 m/s^2, or 32 ft/s^2.
THEREFORE, if you use meters for height, you should use the approximate value of g = 10 m/s^2.
So, in this case a =  = 5 (the numerical value).
ALTERNATIVELY, if you use feet for height, you should use the approximate value of g = 32 ft/s^2.
So, in this case a = = 16 (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).
That's all you need to know.
After my explanations, you can solve the problem on YOUR OWN now,
but for it, you should have your governing equation written correctly.
---------------
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.
================
If this introduction is helpful to you, I will be happy.
If it is not enough for you to solve the problem, come again,
but with one indispensable condition: your equation MUST be written correctly (showing your understanding).
Come again, if you need.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
I'll use x in place of t
And y in place of s(t)
Given equation:
y = -4x^2 + 38x + 3
Your teacher must have mixed up feet with meters because this parabolic projectile equation only works with meters (since half g = 9.8 is 4.9 and it looks like s/he is rounding down to the nearest integer for some reason)
The more accurate projectile equation, when dealing with meters only, is:
y = -4.9x^2 + 38x + 3
where 38 is the initial velocity and 3 is the starting height of the ball
Anyway, I'll ignore the physical real world implications since of course this is just a hypothetical problem to get better practice with parabolas.
----------------------------------------------------------
Part (a)
Plug in x = 1 to find its y value after 1 second.
y = -4x^2 + 38x + 3
y = -4(1)^2 + 38(1) + 3
y = 37
You have the correct answer.
Nice work.
----------------------------------------------------------
Part (b)
For anything of the form
y = ax^2 + bx + c
the vertex (h,k) has its x coordinate defined as
h = -b/(2a)
The k value is determined based on h
From y = -4x^2 + 38x + 3, we have a = -4, b = 38 and c = 3
Let's find h
h = -b/(2a)
h = -38/(2(-4))
h = 4.75
The ball reaches its highest point at 4.75 seconds.
Plug this as the x value into the original equation to find the paired y value of the vertex.
y = -4x^2 + 38x + 3
y = -4(4.75)^2 + 38(4.75) + 3
y = 93.25
The ball reaches the highest point of 93.25 feet.
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