of 24

• date post

01-Feb-2018
• Category

## Documents

• view

216
• download

1

Embed Size (px)

### Transcript of Graphing Quadratic Functions: Vertical motion under ... ‚ Graphing Quadratic Functions:...

• 5.1.1 Graphing Quadratic Functions: Vertical motion under gravity

What goes up, must come down, is a common expression that can be represented by a quadratic equation! If you were to plot the height of a ball tossed vertically, its height in time would follow a simple quadratic formula in time given by the general equation:

2

01( )2

H t h Vt gt

where h0 is the initial height of the ball in meters, V is the initial speed in meters/second, and g is the acceleration of gravity in meters/second2. It is a general equation because it works, not only on Earth, but also on nearly all other astronomical bodies, except for black holes! For black holes, the geometry of space is so distorted that t, V and h0 are altered in complex ways.

For the following problems: A) write the equation in Standard Form, B) determine the coordinates of the vertex of the parabola where H(t) is a maximum; C) determine the axis of symmetry; D) On a common graph for all three problems, draw the parabola for each problem by plotting two additional points using the property of the axis of symmetry, for all positive times during which H(t) > 0 Problem 1 On Earth, the acceleration of gravity is g = 10 meters/sec2. The ball was thrown vertically at an initial speed of V=20 meters/sec (45 mph) from a height of h0 = 2 meters. Problem 2 On Mars, the acceleration of gravity is g = 4 meters/sec2. The ball was thrown vertically at an initial speed of V=20 meters/sec (45 mph) from a height of h0 = 2 meters. Problem 3 On the Moon, the acceleration of gravity is g = 2 meters/sec2. The ball was thrown vertically at an initial speed of V=20 meters/sec (45 mph) from a height of h0 = 2 meters.

Space Math http://spacemath.gsfc.nasa.gov

• Answer Key 5.1.1 Problem 1 Answer: Standard Form is H(t) = at2 + bt +c

A) H(t) = -5t2 + 20t + 2 B) (t,H) = (-b/2a, c - b2/4a) so t= 2 seconds and H(2) = 22 meters. Vertex: (+2,+22) C) The line t = 2 D) See graph below Problem 2 Answer: Standard Form is H(t) = at2 + bt +c A) H(t) = -2t2 + 20t + 2 B) (t,H) = (-b/2a, c - b2/4a) so t=5 seconds and H(5) = 52 meters. Vertex (+5,+52) C) The line t = 5 D) See graph below Problem 3 Answer: Standard Form is H(t) = at2 + bt +c A) H(t) = -t2 + 20t + 2 B) (t,H) = (-b/2a, c- b2/4a) so t=10 seconds and H(2) = 102 meters. Vertex (+10,+102) C) The line t = 10 D) See graph below Note: This would be a good opportunity to emphasize that 1) the plotted graph is not the trajectory of the ball in 2-dimensionsa common misconception. 2) negative values of H(t) are for elevations below the zero point of the launch altitude are unphysical, as are the portions of the curves for T < 0 'before' the ball was thrown. Sometimes math models of physical phenomena can lead to unphysical solutions over part of the domain/range of interest and have to be interpreted against the physical world to understand what they mean.

-110-90-70-50-30-101030507090

110

0 5 10 15 20 25

Time (seconds)

Hei

ght (

met

ers)

EarthMoonMars

Space Math http://spacemath.gsfc.nasa.gov

• 5.2.1 Solving Quadratic Functions by Factoring: Trajectories

After a meteorite strikes the surface of a planet, the debris fragments follow parabolic trajectories as they fall back to the surface. When the LCROSS impactor struck the moon, debris formed a plume of material that reached a height of 10 kilometers, and returned to the surface surrounding the crater. The size of the debris field surrounding the crater can be estimated by solving a quadratic equation to determine the properties of the average trajectory of the debris. The equation that approximates the average particle trajectory is given by

22( ) 2gH x x xV

LCROSS debris plume seen in the dark Shadow of a lunar crater. (NASA/LRO)

Problem 1 The equation gives the height, H(x) in meters, of an average particle ejected a distance of x from the impact site for which V is the speed of the particles in meters/sec and g is the acceleration of gravity on the surface of the Moon in meters/sec2. Factor this equation to find its two roots, which represent the initial ejection distance from the center of the impact crater, and the final landing distance of the particles from the center of the crater. Problem 2 At what distance from the center of the crater did the debris reach their maximum altitude? Problem 3 What was the maximum altitude of the debris along their trajectory? Problem 4 Solve this parabolic equation for the specific case of the LCROSS ejecta for which V = 200 meters/sec and g = 2 meters/sec2 to determine A) the maximum radius of the debris field around the crater, and B) the maximum height of the debris plume.

Space Math http://spacemath.gsfc.nasa.gov

• Answer Key 5.2.1 Problem 1 The equation gives the height, H(t) in meters, of an average particle

ejected from the impact site for which V is the speed of the particles in meters/sec and g is the acceleration of gravity on the surface of the Moon in meters/sec2. Factor this equation to find its two roots, which represent the initial ejection distance from the center of the impact crater, and the final landing distance of the particles from the center of the crater.

Answer: 22( ) 2gH t x xV

can be factored to get 2( ) 0 1 2gH t x xV

The first factor yields X=0 which represents the starting distance of the debris from the center of the crater. The second factor has a value of zero for x = (2V2)/g Problem 2 At what distance from the center of the crater did the debris reach their maximum altitude? Answer: By symmetry, the maximum height was reached half-way between the

endpoints of the parabola or 2 21 2 0

2V Vxg g

Problem 3 What was the maximum altitude of the debris along their trajectory? Answer: Evaluate H(V2/g) to get

22 2 2

2( ) 2V V g VHg g V g

so 2 2

( )2

V VHg g

Problem 4 Solve this parabolic equation for the specific case of the LCROSS ejecta for which V = 200 meters/sec and g = 2 meters/sec2 to determine A) the maximum radius of the debris field around the crater, and B) the maximum height of the debris plume. Answer: A) Maximum radius of ejecta field x = 2V2/g = 2 (200)2/2 = 40,000 meters or 40 kilometers. B) Maximum height of debris plume H=V2/2g = (200)2/4 = 40000/4 = 10,000 meters or 10 kilometers.

Space Math http://spacemath.gsfc.nasa.gov

• 5.3.1 Solving Quadratic Functions with Square-Roots: Speed of sound

Spectacular photo of a fighter jet breaking the 'sound barrier' at a speed of 741 mph or 331 meters/sec (Courtesy Ensign John Gay).

The speed of average air molecules (mostly nitrogen and oxygen) is related to the atmospheric temperature by the formula:

2 400 V T

where V is the speed in meters/sec and T is the gas temperature on the Kelvin scale. This speed is important to know because it defines the speed of sound.

You can convert a temperature in degrees Fahrenheit (F) or Celsius (C) to Kelvins (T) by using the formulae:

5 ( 32) 2739

T F or T C 273

Problem 1 - On the coldest day in Vostok, Antarctica on July 21, 1983, the temperature was recorded as F = -128o Fahrenheit; A) What was the temperature ,T, on the Kelvin scale?; B) What was the speed of the air molecules at this temperature in meters/second? Problem 2 - At normal sea-level on an average day in Spring, the temperature is about F = +60o Fahrenheit. What is the speed of sound at sea-level in meters/sec? Problem 3 - The hottest recorded temperature on Earth was measured at El Azizia, Libya and reached +80o Celsius; A) What was this temperature, T, in Kelvins ? B) What was the speed of sound at this location in meters/sec?

Space Math http://spacemath.gsfc.nasa.gov

• Answer Key 5.3.1 Problem 1 - On the coldest day in Vostok, Antarctica on July 21, 1983, the

temperature was recorded as F = -128o Fahrenheit; A) What was the temperature ,T, on the Kelvin scale?; B) What was the speed of the air molecules at this temperature in meters/second? Answer: A) T = 5/9 (-128 - 32) + 273 so T = 184 Kelvin. B) V2 = 400 T so V2 = 400 (184) V2 = 73,600 And taking the square-root of both sides we get V = 271 meters/sec. Problem 2 - At normal sea-level on an average day in Spring, the temperature is about F = +60o Fahrenheit. What is the speed of sound at sea-level in meters/sec? Answer: A) T = 5/9 (60 - 32) + 273 so T = 288 Kelvin. B) V2 = 400 T so V2 = 400 (288) V2 = 115,200 And taking the square-root of both sides we get V = 339 meters/sec. Problem 3 - The hottest recorded temperature on Earth was measured at El Azizia, Libya and reached +80o Centigrade; A) What was this temperature, T, in Kelvin degrees? B) What was the speed of sound at this location in meters/sec? Answer: A) T = (80) + 273 so T = 353 Kelvin. B) V2 = 400 T so V2 = 400 (353) V2 = 141,200 And taking the square-root of both sides we get V = 376 meters/sec.

Space Math http://spacemath.gsfc.nasa.gov

• Solving Quadratic Functions with Square-Roots: Gravitational collapse 5.3.2

This dark cloud, called Thackeray's Gl