<p>Questions 1-2 refer to the following information.</p>
<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>(^2=squared)</p>
<p>1) If the ball reaches its maximum height in k seconds, what is the value of k?</p>
<p>a) 4 b) 8 c) 12 d) 16 e) 24</p>
<p>2) What is the maximum height, in feet, that the ball will reach?</p>
<p>a) 360 b) 370 c) 384 d) 1024 e) 1200</p>
<p>Thanks :)</p>
<p>1) B
2) D</p>
<p>Are my answers correct ??(I did them without a calculator)
My explanation:</p>
<p>1)The curve of the h=graph is a symmetrical inverted U shape thus to find the time of max height I need to find the time that it stops increasing in height. t=0 or t=16 are the times that it is on ground thus height is 0. Since it is symmetrical the time that height stops increasing is in middle of 0 and 16, which is 8.</p>
<p>2) Just put t=8 into the equation h=graph and TADA the answer is D</p>
<p>You are welcome anytime :D:D:D:D
Btw how can I write a sad face and how can I post a link ???</p>
<p>(1) t=-b/(2a) = -256/(-32) = 8, choice (B)</p>
<p>Note: The graph of a quadratic equation y=ax^2+bx+c is a parabola whose max or min occurs at x=-b/(2a) (If a>0 it’s a min, if a<0 it’s a max).</p>
<p>Ok thanks guys…but I still don’t really get the first part because I am not too good with parabolas…and yes, B and D are correct, according to the answer key.</p>
<p>Eliza, the equation of the height of the ball is that of a parabola (inverted U). The first question essentially asks you to find the time or x coordinate of the peak of the inverted U.</p>
<p>Dr. Steve’s method is probably the fastest way to get the answer. His method requires you to understand some properties of a parabola, including the x coordinate of the vertex of a parabola.</p>
<p>Reality’s method is correct and simple. </p>
<p>I think you might want to use a graphing calculator and plot the graph of Y = 256x-16x^2. If you’re using a TI graphing calculator, you should then hit TRACE and the left or right button until you reach the highest Y value on the parabola. Your X value should be 8 at the peak or vertex of the parabola.</p>
<p>Yea I got it already, but thanks :)</p>
<p>Alternatively, if you know calculus…</p>
<p>1) Take the derivative of h(t). Set that equal to 0. Find t.</p>
<p>2) Plug the value of t you found in question 1 back into h(t).</p>
<p>I have no idea what you’re talking about. All I have taken are two New York State Intergrated Algebra Courses.</p>