A Projectile Is Thrown Upward

Question 115783This question is from textbook Algebra For higher students
: A projectile is thrown upward so that the its distance to a higher place the basis later t seconds is h(t)=-16t^2+308t
After how many seconds does it reach its maximum height?
What is that maximum elevation?
The surface area of a square is numerically 12 more than the perimeter. Observe the length of the side?
These two problems are really hard tin anyone help. This question is from textbook Algebra For higher students

Found 3 solutions by bucky, amalm06, ikleyn:

Respond past bucky(2189) (Show Source):
Y'all tin 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 equally:
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and the start question you are asked is to notice the time that the object reaches its maximum
height. The object starts out at footing level, rises to a peak, and then falls dorsum to
footing level. Neglecting air resistance and other minor considerations, it spends half its
time rising and half its time falling dorsum to ground level. So one way you can find the
time information technology takes to reach its elevation height is to find the fourth dimension of launch and the total time that
goes by until it hits the ground ... then separate that time by two.
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Notice that at ground level the summit given by h(t) is zero. So permit's substitute goose egg
into the equation for h(t) and become (after reversing the sides of the equation to get information technology
into a fiddling more familiar course):
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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 go:
.

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Notice that this equation will exist true if either of the factors is equal to zero because a
multiplication by zero on the left side volition make the left side equal the zilch on the right
side.
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Setting the beginning cistron [which is t] equal to zero results in:
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This means that at t = 0 seconds the object is at basis launch. No surprise here.
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Side by side setting the second factor [which is -16t + 308] equal to zero gives:
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Solve this by commencement subtracting 308 from both sides to go:
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and so dividing both sides by -xvi 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 one-half of that time was spent falling back downwards,
the fourth dimension at which the object reaches its acme is . So at 9.625 seconds
later on launch the object is at its maximum height.
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Some other style you lot can notice this time is to employ part of the quadratic formula. Recollect that
the quadratic formula applies to quadratic equations of the form:
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If you lot compare this with your height equation you will see that a = -sixteen, b = 308, and c = 0.
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Then but employ that portion of the quadratic formula that is to notice 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, but a little different mode of getting it.
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Then to observe the maximum superlative, but substitute 9.625 seconds for t in the summit equation
to get:
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And so the object rises to a height of 1482.25 anxiety. That's quite a throw!!!! Check your problem
to come across if the 308 is the correct multiplier of the t term. If it is, that'southward really the
reply .... 1482.25 feet upward and 1482.25 feet back downwards again.
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Next problem.
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The Area of a foursquare that has S as the length of 1 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 yous take 12 away from the Area, the outcome
volition equal the Perimeter. In equation form this is:
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Go this into the standard quadratic form past subtracting 4S from both sides to become:
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This equation factors to:
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As in the previous trouble, this equation will be truthful if either factor equals zero.
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Setting the commencement factor equal to cipher results in:
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Subtract 2 from both sides and you get:
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This answer doesn't brand sense ... a side of minus ii length??? Ignore information technology.
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Setting the second factor equal to zero gives:
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Add together 6 to both sides and you have:
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This looks better. A square with a side of 6 has an expanse of 6^2 = 36 and a perimeter of
half dozen*4 = 24. The expanse is 12 more than the perimeter, only as the problem specified. So the
side length you were to find is Southward = 6.
.
Hope this helps with your understanding of these two problems. It's pretty tardily so you had
better use your calculator to check my math. The process is correct, but I may accept permit
a calculation error skid in ... I don't think so but better condom than deplorable ...
.

Reply by amalm06(224) (Show Source):
You tin put this solution on YOUR website!
If the position is denoted every bit h(t), then the instantaneous velocity at any given time is just the first derivative, h'(t), while the acceleration is the second derivative, h''(t).
If y'all differentiate the given position function twice, y'all will obtain a value of acceleration equal to -32 ft/sec^2, where I take called the units arbitrarily. The gravitational forcefulness (mg) acts on the projectile equally it moves upward. Ignoring air resistance, this is the only force interim on the projectile. Since the acceleration vector points radially inwards, the projectile slows downward equally it rises. At the highest point, all of the kinetic energy of the projectile becomes potential energy, east.1000. KE=PE.
To find the velocity of the projectile at it's highest point, take the kickoff derivative:
h'(t)=-32t+208
Set up h'(t) equal to 0: -32t+208=0
t=ix.625 s (Respond)
The projectile reaches its maximum height later nine.625 s
The maximum height is given as follows:
h(vi.5)= -16(9.625)(nine.625)+308(9.625)=-1482.25+2964.5=1482.25 ft (Answer)
Denote the side length of a square by s
So the total perimeter P of the square is 4s
The total surface area A of the foursquare is s^two
We are told that the area is 12 more than than the perimeter.
Therefore, s^ii=4s+12
Rewrite equally a quadratic equation:
s^2-4s-12=0
Cistron:
(s+two)(s-six)=0
Only the positive root is meaning. Therefore, s=6 (Reply)

Answer by ikleyn(46021) (Prove Source):
You tin put this solution on YOUR website!
.
Allow me explain y'all EVERYTHING about these bug,
from the very showtime to as far as you need to know it Now.
1. You may often meet these problems on a projectile thrown vertically upwards. The equation for the elevation over the ground normally has ONE OF TWO POSSIBLE forms: a) h(t) = -sixteen*t^two + v*t + c In this course, the equation is written for the tiptop h(t) over the ground measured in anxiety. The value of "16" is the half of the value of the gravity acceleration g = 32 ft/s^2. The sign "-" at the outset term ways that the gravity dispatch is directed down, while the "y"-centrality of the coordinate arrangement is directed vertically up, in the contrary direction. The value of "v" in this equation is the value of the initial vertical velocity. The value of "c" is the initial top over the ground. The ground level is assumed to be 0 (nada, Nil). In other words, the origin of the coordinate system is at the ground. b) h(t) = -v*t^2 + v*t + c It is some other form of the "height" equation for the same process. In this class, the height h(t) is measured in meters (instead of feet). The value of "5" at the offset term is the same gravity dispatch, simply this time expressed in "yard/s^ii" units" thou = 10 m/s^2. Actually, more than precise value is g = ix.8 m/s^ii, therefore, sometimes, this equation goes with the commencement term -4.9. The value of "5" is the vertical velocity, expressed in m/due south. The value of "c" is the initial summit over the ground in meters. 2. In any case, when such issues come up from Algebra (equally Algebra issues), they are treated in THIS Fashion: The question "discover the maximal height" is the aforementioned equally "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 than usual "x" variable. Next, when the question is about the maximum/minimum of a quadratic form q(10) = 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. three. Same problems may come from CALCULUS. In Calculus, they are treated in this manner: to find the maximum (minimum), take the derivative and equate it to nil. It will give the equation to find "t". 4. Same problems may come up from PHYSICS. In Physics, they are treated in this way: the maximum height is achieved when the verical velocity becomes equal to zippo. It will give the equation to discover "t": t = . 5. The amazing fact is that different approaches from different branches of Math and Science give the same answer. Astonishing ? - Yep, of cource, without doubts for young students. Amazing ? - Yep, merely not then much for more mature students, who understand that all these branches of noesis study the same Nature's phenomenos. And then, the results should (and must) be identical.

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On finding maximum/minimum of a quadratic function run across my lessons
- HOW TO consummate the foursquare to observe the minimum/maximum of a quadratic role
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO consummate 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 upward and down
- Problem on an pointer shot vertically up
- Problem on a ball thrown vertically up from the acme of a tower
- Trouble 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 role 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://world wide web.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and apply it when it is needed.