Question 1183278: I need help with this please. A baseball player throws a baseball from a height of 1 meter above the ground and its height is given by the equation 𝒉 = −𝟑. 𝟐𝒕^𝟐 + 𝟏𝟐.𝟖𝒕 + 𝟏, where 𝒉 is the height in metres above the ground, and t, in seconds, is its
time in the air [see the image below]. When, to the nearest tenth of a second, will the ball hit the ground? This is the image https://giannepaulaperalta.files.wordpress.com/2018/07/3075613-6448579381.jpg (you could pls copy and paste the link on a new tab). Also responding to the tutor that asked if I created this question on my own. To answer that no I didn't. This is a question from my assignments given to me by my teacher.
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
The basic formula for the height as the function of time is written INCORRECTLY in your post.
It has a FATAL ERROR, which makes the solution NONSENSICAL.
To get the basic knowledge on the subject, read my post to the end.
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The problems on a projectile thrown-shot-launched vertically up are very popular.
But, as I often observed, the students who meet such problems for the first time, often write
the basic equation incorrectly, because they do not understand the meaning of its terms.
Therefore, I wrote this introductory lesson specially for beginners who don't know the subject AT ALL.
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,
if you use meters for height.
So, in this case, when you use meters as the measure of height, a =  =  m/s^2 or 5 m/s^2 (approximate numerical value).
+--------------------------------------------------------------+
| It can not be 3.2 m/s^2, as it is written in your post. |
+--------------------------------------------------------------+
(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 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
- OVERVIEW of lessons on a projectile thrown/shot/launched vertically up
in this site.
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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.
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Regarding your teacher, you may show him (or her) this my post. Or, if you hesitate do it,
give me his (or her) contact, and I will ask him (or her), why he (or she) gives wrong assignments.
In the country where I growth and got my school and university education,
it was UNTHINKABLE that a Math teacher gave an incorrect assignment.
I can not recall even one single such case.
It was not because the teachers were absolutely perfect - they were normal people (very professional, although).
But the educational system itself was so perfect that failed cases were impossible to happen inside of this system.
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