Question 1117207: A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)=116-16t^2 .
What is the maximum height that the ball will reach? Do not round your answer.
Found 2 solutions by Alan3354, ikleyn: Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)=116-16t^2 . What is the maximum height that the ball will reach? Do not round your answer.
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Any value of t --> a smaller value of h(t)
Max = 116
Answer by ikleyn(52754) (Show Source):
You can put this solution on YOUR website! .
A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t)=116-16t^2 .
What is the maximum height that the ball will reach? Do not round your answer.
~~~~~~~~~~~~~~~~~
Hey,
the function written in your post DOES NOT describe the ball thrown vertically up, since the term
responsible for the vertical velocity is ABSENT.
It is YOUR ERROR in writing your post.
Double check you source.
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Let me explain you EVERYTHING about these problems,
from the very beginning to as far as you need to know it NOW.
1. You may often meet these problems on a projectile thrown vertically upward.
The equation for the height over the ground usually has ONE OF TWO POSSIBLE forms:
a) h(t) = -16*t^2 + v*t + c
In this form, the equation is written for the height h(t) over the ground measured in feet.
The value of "16" is the half of the value of the gravity acceleration g = 32 ft/s^2.
The sign "-" at the first term means that the gravity acceleration is directed down,
while the "y"-axis of the coordinate system is directed vertically up, in the opposite direction.
The value of "v" in this equation is the value of the initial vertical velocity.
The value of "c" is the initial height over the ground.
The ground level is assumed to be 0 (zero, ZERO). In other words, the origin of the coordinate system is at the ground.
b) h(t) = -5*t^2 + v*t + c
It is another form of the "height" equation for the same process.
In this form, the height h(t) is measured in meters (instead of feet).
The value of "5" at the first term is the same gravity acceleration, but this time expressed in "m/s^2" units" g = 10 m/s^2.
Actually, more precise value is g = 9.8 m/s^2, therefore, sometimes, this equation goes with the first term -4.9.
The value of "v" is the vertical velocity, expressed in m/s.
The value of "c" is the initial height over the ground in meters.
2. In any case, when such problems comes from Algebra (as Algebra problems), they are treated in THIS WAY:
The question "find the maximal height" is the same as "find the maximum of the quadratic form h(t) = -16t^2 + vt + c.
It doesn't matter that the quadratic function presented as the function of "t" instead of more usual "x" variable.
Next, when the question is about the maximum/minimum of a quadratic form
q(x) = ax^2 + bx + c,
the Algebra teach us that the maximum is achieved at x = .
In your case this value of "t", which provides the maximum height, is t = = 13.75 seconds.
3. Same problems may come from CALCULUS. In Calculus, they are treated in this way:
to find the maximum (minimum), take the derivative and equate it to zero.
It will give the equation to find "t".
4. Same problems may come from PHYSICS. In Physics, they are treated in this way:
the maximum height is achieved when the verical velocity becomes equal to zero.
It will give the equation to find "t": t = .
5. The amazing fact is that different approaches from different branches of Math and Science give the same answer.
AMAZING ? - Yes, of cource, without doubts for young students.
AMAZING ? - Yes, but not so much for more mature students, who understand that all these branches of knowledge
study the same Nature's phenomenos. So, the results should (and must) be identical.
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On finding maximum/minimum of a quadratic function see my lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
in this site.
My lessons on a projectile thrown/shot/launched vertically up are
- 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.
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 "Finding minimum/maximum of quadratic functions"
and 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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