Question 115783This question is from textbook Algebra For college students
: A projectile is thrown upward so that the its distance above the ground after t seconds is h(t)=-16t^2+308t
After how many seconds does it reach its maximum height?
What is that maximum height?
The area of a square is numerically 12 more than the perimeter. Find the length of the side?
These two problems are really hard can anyone help.
This question is from textbook Algebra For college students
Found 3 solutions by bucky, amalm06, ikleyn: Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! First problem:
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You are given the equation of the height of an object that is thrown upward as:
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and the first question you are asked is to find the time that the object reaches its maximum
height. The object starts out at ground level, rises to a peak, and then falls back to
ground level. Neglecting air resistance and other minor considerations, it spends half its
time rising and half its time falling back to ground level. So one way you can find the
time it takes to reach its peak height is to find the time of launch and the total time that
goes by until it hits the ground ... then divide that time by 2.
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Notice that at ground level the height given by h(t) is zero. So let's substitute zero
into the equation for h(t) and get (after reversing the sides of the equation to get it
into a little more familiar form):
.

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Note that t is a common factor of both the terms on the left side, so it can be factored to
make the left side become:
.

.
Notice that this equation will be true if either of the factors is equal to zero because a
multiplication by zero on the left side will make the left side equal the zero on the right
side.
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Setting the first factor [which is t] equal to zero results in:
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This means that at t = 0 seconds the object is at ground launch. No surprise here.
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Next setting the second factor [which is -16t + 308] equal to zero gives:
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Solve this by first subtracting 308 from both sides to get:
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and then dividing both sides by -16 to get:
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This tells us that from time of launch at t = 0, 19.25 seconds later the object hits the ground.
Since half of that time was spent rising and half of that time was spent falling back down,
the time at which the object reaches its peak is . So at 9.625 seconds
after launch the object is at its maximum height.
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Another way you can find this time is to apply part of the quadratic formula. Recall that
the quadratic formula applies to quadratic equations of the form:
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If you compare this with your height equation you will see that a = -16, b = 308, and c = 0.
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Then just use that portion of the quadratic formula that is to find the
time at the peak. Substituting 308 for b and -16 for a results in:
.
seconds.
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This is the same answer, just a little different way of getting it.
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Then to find the maximum height, just substitute 9.625 seconds for t in the height equation
to get:
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So the object rises to a height of 1482.25 feet. That's quite a throw!!!! Check your problem
to see if the 308 is the correct multiplier of the t term. If it is, that's really the
answer .... 1482.25 feet up and 1482.25 feet back down again.
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Next problem.
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The Area of a square that has S as the length of one side is given by the equation:
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and the Perimeter of the square is the sum of the lengths of all its sides:
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Since the Area is 12 more than the Perimeter, if you take 12 away from the Area, the result
will equal the Perimeter. In equation form this is:
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Get this into the standard quadratic form by subtracting 4S from both sides to get:
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This equation factors to:
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As in the previous problem, this equation will be true if either factor equals zero.
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Setting the first factor equal to zero results in:
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Subtract 2 from both sides and you get:
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This answer doesn't make sense ... a side of minus 2 length??? Ignore it.
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Setting the second factor equal to zero gives:
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Add 6 to both sides and you have:
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This looks better. A square with a side of 6 has an area of 6^2 = 36 and a perimeter of
6*4 = 24. The area is 12 more than the perimeter, just as the problem specified. So the
side length you were to find is S = 6.
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Hope this helps with your understanding of these two problems. It's pretty late so you had
better use your calculator to check my math. The process is correct, but I may have let
a calculation error slip in ... I don't think so but better safe than sorry ...
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Answer by amalm06(224) (Show Source):
You can put this solution on YOUR website! If the position is denoted as h(t), then the instantaneous velocity at any given time is simply the first derivative, h'(t), while the acceleration is the second derivative, h''(t).
If you differentiate the given position function twice, you will obtain a value of acceleration equal to -32 ft/sec^2, where I have chosen the units arbitrarily. The gravitational force (mg) acts on the projectile as it moves upwards. Ignoring air resistance, this is the only force acting on the projectile. Since the acceleration vector points radially inward, the projectile slows down as it rises. At the highest point, all of the kinetic energy of the projectile becomes potential energy, e.g. KE=PE.
To find the velocity of the projectile at it's highest point, take the first derivative:
h'(t)=-32t+208
Set h'(t) equal to 0: -32t+208=0
t=9.625 s (Answer)
The projectile reaches its maximum height after 9.625 s
The maximum height is given as follows:
h(6.5)= -16(9.625)(9.625)+308(9.625)=-1482.25+2964.5=1482.25 ft (Answer)
Denote the side length of a square by s
Then the total perimeter P of the square is 4s
The total area A of the square is s^2
We are told that the area is 12 more than the perimeter.
Therefore, s^2=4s+12
Rewrite as a quadratic equation:
s^2-4s-12=0
Factor:
(s+2)(s-6)=0
Only the positive root is significant. Therefore, s=6 (Answer)
Answer by ikleyn(52775) (Show Source):
You can put this solution on YOUR website! .
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 come 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 teaches us that the maximum is achieved at x = .
In your case this value of "t", which provides the maximum height, is t = = 9.625 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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