Transcript THE VERTICAL MOTION MODEL - Santa Rosa County School District

Space Exploration
Quadratic Equations
Vertical Motion
Gravity
The Mathematics Connection
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.
y = ax2 + bx + c
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.
height
time
Let’s do a quick review. . .
• The quadratic equation is
y = ax2 + 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 ± √(b2 – 4ac)
2a
The Quadratic Equation
If the coefficient of the x2
term is positive:
The parabola opens
UP
If the coefficient of the x2
term is negative:
The parabola opens
DOWN
The Vertical Motion Model
How do the equations connect?
y = ax2 + bx + c
h(t) = -16t2 + v0t + c
(if height is in feet)
h(t) = -4.9t2 + v0t + c
(if height is in meters)
Where t is time in seconds
h(t) is height above ground after t seconds
v0 is the initial velocity of the object
c is the initial height of the object.
Where do we get -16t2 and -4.9t2
Gravity pulls objects toward the center of the
earth (“down” to us) at an acceleration of 32
feet per sec2 (English measure) or 9.8 meters per
sec2 (metric measure).
The coefficient of the t2 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) = -16t2 + 10t + 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)2 + 10(0.5) + 3
h(0.5) = -4 +5+ 3
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.
ft
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?
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) = -16t2 + v0t + 42.
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:
h(t) = -16t2 +6t + 42
41 = -16(0.5)2 + v0 (0.5) + 42
41 = -4 + 42 + 0.5v0
3 = 0.5v0
v0 = 6
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) = -16t2 -6t + 42
Let’s recap what we found
h(t) = -16t2 +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) = -16t2 -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).
0 = -16t2 -6t + 42
Use the quadratic
formula to solve.
X = -(-6) ± √[(-6)2 -4(-16)(42)]
2(-16)
X = -1.82 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!
Oh, and of course. . .
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!
GO GATORS!!!!