Dr.Chungs SAT Math Question

[thatoneman]

<p>A ball is thrown straight up from the ground with an initial velocity of 256 feet per second. The equation h=256t-16t^2 describes the height the ball can reach in "t" seconds. </p>

<p>What is the maximum height, in feet, that the ball will reach? </p>

<p>I know what the answer is, I looked in the back of the book, what is your answer and how do you explain it. </p>

<p>a-360
b-370
c-384
d-1024
e-1200</p>

[IceQube]

<p>Complete the square. The answer is D, 1024.</p>

<p>The vertex of a parabola has special properties. It can give you the minimum height (if the parabola is shaped like a U) or it can give you the maximum height when the parabola is a upside down U (as in this case). </p>

<p>How do you find the vertex? You must convert the equation Chung gave you to the so-called “graphing” form … ** y = a(x-h)^2 + k **</p>

<p>To do this, you must complete the square.</p>

[Sikorsky]

<p>Another approach:</p>

<p>The graph of the equation h=256t-16t^2 is a parabola that opens downward. The dependent variable is height (h), so it goes where y normally goes, and the independent variable is time (t), so it goes where x normally goes.</p>

<p>The greatest height attained is at the vertex of the parabola.</p>

<p>So graph the equation y=256x-16x^2 on your calculator, and use the CALC function to calculate the vertex. The y-coordinate of the vertex is the maximum height.</p>

<p>(x-post with Ice Qube’s revisions above.)</p>

[realityisadream]

<p>This is not a SAT question, it is CALCULUS based. It can be solved by differentiation.</p>

[Sikorsky]

<p>

sorry so i was creeping on this thread trying to figure it out… so after complete square do u get h= -16(t-8)^2 +1024. and 1024 is the answer?

</p>

<p>When you complete the square and rewrite the function in the form y = a(x-h)^2 + k, the coordinates of the vertex are (h,k).</p>

<p>So, in this case, the vertex of the parabola is (8,1024). The y-coordinate is max height (1024 ft.), and the time when it occurs is the x-coordinate (i.e., 8 sec.).</p>

[IceQube]

<p>This type of question may appear on an actual SAT since it is a standard Algebra II problem.</p>

<p>Here is how to complete the square:</p>

<p><a href=“http://i.min.us/iddMrSUo6.JPG[/url]”>http://i.min.us/iddMrSUo6.JPG&lt;/a&gt;&lt;/p&gt;

[realityisadream]

<p>

@xcbballuver

will a question like this actually be on the SAT? i havent seen any questions like this one on the practice tests.

</p>

<p>It is ALMOST IMPOSSIBLE for a question like this to appear on the SAT,but it can appear on the SAT Math 2 subject test.</p>

[Sikorsky]

<p>

This is not a SAT question, it is CALCULUS based. It can be solved by integration.

</p>

<p>You mean by differentiation?</p>

<p>You can do this problem with calculus, but clearly you don’t need calculus to do it.</p>

<p>This would be a fair question by some time in May in my Algebra I class.</p>

[realityisadream]

<p>^ Yup differentiation
Sorry for the mistake.</p>

[kestrel24]

<p>Here’s how I got the answer. Rearrange the equation so that it looks like f(x)=ax^2+bx+c.</p>

<p>h=-16t^2+256t</p>

<p>You should know that -b/2a will give you the x coordinate of the vertex. Plug in the numbers and you get 8. Since you need the y coordinate, substitute 8 back into the equation, giving h=1024 (D).</p>

[Sikorsky]

<p>Kestrel’s way is what I would actually expect most of my algebra students to do, but you can also complete the square or find the answer graphically.</p>

<p>(Or, of course, you can use first-semester calculus if you know it.)</p>

[AstroBlue]

<p>I guess you could use calculus…Our you could use the fact that the x value for hte vertex of a parabola is -b/2a</p>

[xiggi]

<p>

It is ALMOST IMPOSSIBLE for a question like this to appear on the SAT,but it can appear on the SAT Math 2 subject test.

</p>

<p>Is this that different?</p>

<p>

At time t=0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet, after t seconds was given by the function h above, in which c and d are positive constants. If the ball reached its maximum height of 106 feet at time t=2.5, what was the height, in feet, of the ball at time t=1?

</p>

<p>Appeared in 2005 BB1. </p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/159048-function-question.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/159048-function-question.html&lt;/a&gt;&lt;/p&gt;

[realityisadream]

<p>^OHHHHHHH. Thank you very very much for the clarification.
I will be prepared if it came in the SAT now. I really am grateful.</p>

[thatoneman]

<p>Guys, the book says C. :(</p>

[nothingto]

<p>^ Lol, dun dun dun dun</p>

[thatoneman]

<p>I know. I thought it was 1024 when I solved it but back of the book was like C. I was like “I’m stupid…” Then I realized it might be a typo. I hope it is… if any of you all have Dr.Chungs green book can you verify it?</p>

[nothingto]

<p>^What question is it/page?</p>

[herozero1234]

<p>My method was simple, although there are 2 different ways to easily do it.</p>

<p>The first, completing the square, but i didnt do it like this.</p>

<p>the next, you can just do -b/2a , -250/2(-16)=8…then plug 8 into the equation and you get 1024.</p>

<p>Or else you can just plug each of the options into the equations, it might take more time, but it works too.</p>

[DrSteve]

<p>Here’s a quick solution using calculus:</p>

<p>The derivative of h is h’ = 256 - 32t.
Setting h’=0 yields 256 - 32t = 0. So 32t=256, and t=256/32=8.</p>

<p>Finally plug t=8 into the original function:
h=256*8-16(8)^2=1024</p>