![]() Let's examine those points a little bit more closely. The last important point is the location on the graph that describes when the projectile reaches the ground. The second point is the maximum height of the projectile. Here is the first important point on the graph, which is the initial height of the projectile. Each graphing calculator is a little different. Those points can be easily located using various features of a graphing calculator, like adjusting its "window," using "zoom::fit," "trace," and other commands. There are three important locations on the graph. ![]() So, think of each point on the graph as (time, vertical height), not (x,y), and the rest of this lesson will be much more clear. Keep in mind that the horizontal dimension (→) represents time and the vertical dimension (↑) represents the vertical height of the projectile. Here is what the graph of our specific polynomial function looks like. There are also several online graphing calculators that can graph. There are several great graphing calculators available from Casio and Texas Instruments. Graphing it can be easily accomplished with the use of a graphing calculator. The next steps involve graphing the polynomial function. So, our specific polynomial function is this. The 50 is the initial vertical distance (height) of the projectile, 'd.' The 100 is the initial vertical velocity, 'v.' Since we are on Earth, the gravitational force is 32. The first step is to identify the values in the problem. If a projectile has an initial height of 50 feet and it's given an initial upward velocity of 100 feet/second, write a formula that describes the height of the projectile over time and determine all of its critical points. Let's look at an example problem and see how we use it to make a specific function. In the following two sections, a graphing calculator will be used to locate the maximum height of projectiles and the time at which they return to the ground. Ideo: Deriving the Formula for the Vertex of a Quadratic Function If you already know how to complete the square and understand translations ( transformations), then skip the lessons and go straight to the video. (b) The horizontal motion is simple, because a x 0 and v x is thus constant. Caution: It is suggested that you first watch the two prerequisite lessons so that you will understand the video. Figure 3.35 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. If you are interested in learning of the origins of the algebra formula for finding the time value when the projectile reaches its maximum height (the x-value of the vertex of a parabola), watch the video marked with the 'v,' immediately below. In this section you will learn how to find that height and time using algebra. ![]() yo 0, and, when the projectile is at the maximum. At this point, all we have to do is plug t 1.27s into Equation 13A.1 and evaluate: x V0xt 9.97m s (1.27s) 13m. There are several methods for finding the maximum height of a projectile and the time it gets there. The maximum height, ymax, can be found from the equation: vy 2 voy2 + 2 ay (y - yo). Note that the whole time it has been moving up and down, the projectile has been moving forward in accord with Equation 13A.1, x V0xt.
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