Math Learning Center Boise State 2010, Quadratic Modeling STEM 10

Transcription

1 Quadratic Modeling STEM 10 Today we are going to put together an understanding of the two physics equations we have been using. Distance: Height : Recall the variables: o acceleration o gravitation force (be careful of direction) o height at time t o height at time 0 o time o position at time t o position at time 0 Recall: Typically in physics distances are measured in meters per second. Below is the conversion from miles per hour: 1 meter per second = miles per hour Thus 30 m/s is just under 70mph. First let s consider that a baseball player throws the ball straight up into the air at a velocity of 30 m/s. The initial height off the ground when the ball leaves the hand is 2 m. And recall that gravitational acceleration is given as 9.8 m/s 2. Write the equation. (Be careful, gravity is 9.8 m/s 2 but this is in the downward direction. In the equation is this positive 9.8 or negative 9.8?) Use the t-chart to fill in some values to assist in graphing the equation. h t

2 Discussion: Discuss what is happening in this problem and include where the ball hits the ground. Recall the quadratic formula So where did the ball hit the ground? The answer should be at the baseball player s feet. The height equation is only used for height and for vertical movement. On the other hand, the distance equation is used for finding horizontal distance. Thus, if we consider what happens when an outfielder is throwing the ball, trying to make a play at home. The ball is thrown slightly upward and towards home base at a 15 degree angle at 40m/s. How far will the ball travel before it hits the ground? In the problem we have used the direction 15 degrees. This implies that some of the velocity is upward (height) and some of the distance is horizontal (distance). In order to complete this problem, we need to determine the amount of the velocity that is vertical and the amount of the velocity that is horizontal. To determine these values, we use a form of math called trigonometry (or shorthand trig ). Thus, we will break down the velocity into direction for you, knowing you will have this opportunity later. The vertical velocity (height equation) is m/s and the horizontal velocity (distance equation) is m/s. Realize at this point we are using in both equations to designate the initial velocity in the desired direction. It is up to the reader to determine the context and use the correct value. In solving this problem, we need to first find the total time the ball is in the air. This is done by using the height equation. Set up the height equation with. (Should the sign for gravity should be positive or negative?)

3 Solve the height equation for when the ball hits the ground by letting Be careful of your positive and negative signs. Also ignore any answers outside of the implied domain (feasible region). Now that we know how long the ball is in the air, we can plug this time into the distance formula to find out how far the ball will travel. In this case and assuming that wind resistance is negligible, we can let Thus the ball traveled: Your answer is in terms of meter. We can convert meters to feet by multiplying by This should give an answer of approximately 300 feet which would represent a good throw in our major league ball parks. The combination of the height equation and the distance equation gives us the opportunity to study the motion of any projectile. The student needs to understand that anything that flies is a projectile. With these two equations students will be able to compute the height and distance of any projectile that is not supported with wings or other devices. In those cases, these equations are the basis for more complicated mathematical analysis that will be learned later. In this set of activities we will remain with sports as it is easy to visualize. For the next set of problems, consider that we have a shot-putter throwing at a velocity of 20 m/s. Again assume (positive or negative?). Of the following angles, at which angle would the shot-putter get the farthest distance? (We have computed vertical velocity and horizontal velocity.) 1) 10 degrees: Vertical velocity 3.47 m/s. Horizontal velocity m/s.

4 2) 20 degrees: Vertical velocity 6.84 m/s. Horizontal velocity m/s. 3) 30 degrees: Vertical velocity 10 m/s. Horizontal velocity m/s. 4) 40 degrees: Vertical velocity m/s. Horizontal velocity m/s. So far our primary concern of the quadratic equation is to find zeros. Let s look at some problems with projectiles concerned with finding the vertex (or maximum height). 5) If a projectile is shot from ground level at 50m/s, what is its maximum height and when will it land? 6) If the same projectile is shot from a building rooftop (80 m above the ground) under the same conditions, what is its maximum height and when will it land at ground level?

5 7) If a projectile is shot from a rooftop 40 m above the ground at a velocity of 25 m/s, will the 8) If a projectile is shot from a rooftop 40 m above the ground at a velocity of 30 m/s, will the 9) If a projectile is shot from a rooftop 40 m above the ground at a velocity of 40 m/s, will the