Presentation on theme: "Quadratic Equations Gravity Vertical Motion The Mathematics Connection Space Exploration."— Presentation transcript:
Quadratic Equations Gravity Vertical Motion The Mathematics Connection Space Exploration
THE VERTICAL MOTION MODEL We learned in the NASA video that the aircraft flew in a parabolic arc. That arc is described by the quadratic equation. We also use a quadratic equation to describe how objects behave when gravity is involved. In the VERTICAL MOTION MODEL, the height of the object above the ground is a function of the amount of time it has been in motion. y = ax 2 + bx + c height time
Let’s do a quick review... The quadratic equation is y = ax 2 + bx + c The graph of a quadratic is a parabola If you want to solve the equation for x, you can use the quadratic formula x = -b ± √(b 2 – 4ac) 2a
The Quadratic Equation If the coefficient of the x 2 term is positive: If the coefficient of the x 2 term is negative: The parabola opens DOWN The parabola opens UP
The Vertical Motion Model How do the equations connect? y = ax 2 + bx + c h(t) = -16t 2 + v 0 t + c (if height is in feet) h(t) = -4.9t 2 + v 0 t + c (if height is in meters) Where t is time in seconds h(t) is height above ground after t seconds v 0 is the initial velocity of the object c is the initial height of the object.
Where do we get -16t 2 and -4.9t 2 Gravity pulls objects toward the center of the earth (“down” to us) at an acceleration of 32 feet per sec 2 (English measure) or 9.8 meters per sec 2 (metric measure). The coefficient of the t 2 term is acceleration. So, why are the coefficients -16 and -4.9? Because the AVERAGE acceleration per second is what is used. An object’s velocity will be greater at the end of the one-second interval than at the beginning of the interval. If acceleration at t=0 is 0 and at t=1 is -9.8, then the average is -4.9 in that interval. The negative sign is because the acceleration is down (negative)
One more reminder Velocity is speed with direction. The sign of the coefficient of the t term will be positive or negative, depending upon the direction of the initial speed.
Let’s test this out 1. Give a verbal description of the vertical motion modeled by h(t) = -16t t + 3 A tennis player hit a ball from a height of 3 feet above the ground at an initial velocity of 10 feet per second. 2. How high off the ground will the tennis ball be at 0.5 seconds? h(0.5) = -16(0.5) (0.5) + 3 h(0.5) = h(0.5) = 4 feet above the ground
Investigation Mario is a carpenter on a high-rise construction site. He uses the open elevator to go from floor to floor. Two times this week, Mario dropped his hammer while riding the elevator. The graphs of the two incidents are on the right. What things can you tell immediately about the two mishaps? Can you describe any of the coefficient values? ft sec
Here are some possibilities Both graphs start at 42 feet above the ground, so both times, Mario dropped the hammer from 42 feet. That means that c = 42 The blue graph starts upward. Mario was ascending in that mishap. That means the middle coefficient is positive The red graph starts decreasing immediately. The elevator was standing still or descending. If it was descending, the middle coefficient is negative. The first coefficient is -16. ft sec
What we now know: So far, we have h(t) = -16t 2 + v 0 t How can we find the value of the initial velocity for each graph? Let’s start with the blue graph. Plug the values of the coordinates of one of the points into the equation: 41 = -16(0.5) 2 + v 0 (0.5) = v 0 3 = 0.5v 0 v 0 = 6 h(t) = -16t 2 +6t + 42 Mario was 42 feet above the ground, traveling up at 6 ft per second.
Now, you determine the equation of the red graph You can use either set of coordinates h(t) = -16t 2 -6t + 42
Let’s recap what we found h(t) = -16t 2 +6t + 42 Blue graph shows that Mario was going up on the elevator at a velocity of +6 when he dropped hammer from a height of 42 feet. h(t) = -16t 2 -6t + 42 Red graph shows that Mario was going down on the elevator at a velocity of -6 when he dropped hammer from a height of 42 feet.
What if a spider was riding on the hammer? When Mario dropped the hammer while he was descending a spider went along for the ride. How long did the spider experience microgravity before the hammer hit the ground? Ask yourself – the hammer was how high above the ground when it hit the ground? Plug that value into the equation for h(t) and use the quadratic formula to solve for t.
Calculate!** How high was the hammer above the ground when it hit the ground? 0 feet Plug that into the equation For h(t). Use the quadratic formula to solve. 0 = -16t 2 -6t + 42 X = -(-6) ± √[(-6) 2 -4(-16)(42)] 2(-16) X = x = 1.44 Since this is a real world problem We use the positive value. The spider was “space-spider” for 1.44 seconds ** No spiders were injured in this problem
What have you learned? Take a moment to talk with your fellow classmates about today’s lesson: 1. How does microgravity explain why the astronauts appear to be “floating” inside the space shuttle? 2. What do each of the coefficients in the quadratic equation and the vertical motion model represent? 3. Where on the parabolic arc do the passengers on the parabolic flight feel weightless? 4. If you know the equation of a vertical motion problem, how can you tell how high an object will be at.6 seconds?
Wrap it up! We hope that you have learned something from Math Day 2009! We also hope that you enjoyed the journey on the Vomit Comet. The residents of Math Land hope that they both entertained you and helped you learn about microgravity. Your teacher has an extension problem for you to work on. Let us know how you solve it! Oh, and of course...