- NASA LAUNCHES A ROCKET 
AT T = 0 SECONDS.
IT'S HEIGHT IN METERS 
ABOVE SEA LEVEL
AS A FUNCTION OF TIME IS GIVEN 
BY THE FUNCTION H OF T.
WE WILL ASSUME THE ROCKET WILL 
SPLASH DOWN INTO THE OCEAN.
WE WANT TO ANSWER 
TWO QUESTIONS.
1. WHAT TIME WILL THE ROCKET 
SPLASH INTO THE OCEAN?
2. HOW HIGH ABOVE SEA LEVEL 
WILL THE ROCKET REACH?
NOW, WE JUST SOLVED 
THIS PROBLEM
USING A GRAPHING CALCULATOR,
BUT NOW WE'RE GOING 
TO SOLVE IT
USING WHAT WE KNOW 
ABOUT QUADRATIC FUNCTIONS
AND QUADRATIC EQUATIONS.
SO BEFORE WE SOLVE THIS
I THINK IT'S GOING TO BE 
HELPFUL TO UNDERSTAND
WHAT'S HAPPENING 
IF WE GRAPH THIS FUNCTION.
SO I'VE ALREADY GRAPHED THIS 
USING SOME GRAPHING SOFTWARE.
SO OUR X AXIS REPRESENTS 
TIME IN SECONDS
AND OUR Y AXIS REPRESENTS 
THE METERS ABOVE SEA LEVEL.
NOTICE AT TIME T = 0 THE 
ROCKET WOULD BE RIGHT HERE,
WHICH WOULD BE 
THE Y INTERCEPT.
THIS REPRESENTS THE INITIAL 
HEIGHT OF THE ROCKET
AND THEN AS TIME PASSES 
THE ROCKET GOES UP,
REACHES A MAXIMUM HEIGHT 
HERE AT THE VERTEX
AND THEN COMES BACK DOWN 
TO THE EARTH.
AND WHEN THE HEIGHT REACHES 0
THAT WOULD BE WHEN THE ROCKET 
SPLASHES INTO THE OCEAN,
WHICH WOULD BE THIS POINT HERE
WHICH WOULD BE 
THE X INTERCEPT.
SO TO FIND HOW LONG IT TAKES 
FOR THE ROCKET
TO SPLASH INTO THE OCEAN WE 
WANT TO FIND THE X INTERCEPT.
AND TO FIND THE MAXIMUM HEIGHT
WE WANT TO FIND THE Y 
COORDINATES OF THE VERTEX.
SO LET'S GO BACK AND DO THAT.
AGAIN, TO FIND WHEN THE ROCKET 
SPLASHES INTO THE OCEAN
WE KNOW THE HEIGHT 
ABOVE SEA LEVEL,
OR H OF T, WOULD BE = TO 0.
SO WE WANT TO SOLVE 
THE EQUATION
0 = -4.9 T SQUARED + 310T 
+ 332.
WELL, THIS IS NOT A FACTORABLE 
QUADRATIC EQUATION,
WHICH MEANS WE'LL HAVE 
TO SOLVE IT
USING THE QUADRATIC FORMULA 
GIVEN HERE BELOW.
SO FOR THE FIRST STEP 
WE NEED TO RECOGNIZE
THAT A IS = TO -4.9, 
B IS = TO 310,
AND C IS = TO 332.
AGAIN, THIS IS BASED 
UPON THE FACT
THAT WE'RE USING THE FUNCTION 
H OF T = AT SQUARED + BT + C.
SO NOW WE'LL SUB THESE 
INTO THE QUADRATIC FORMULA
AND THEN EVALUATE.
SO INSTEAD OF X = WE'LL HAVE
T = -310 + OR - THE SQUARE 
ROOT OF B SQUARED
OR 310SQUARED - 4 x A, 
WHICH IS -4.9 x C,
WHICH IS 332 DIVIDED 
BY 2 x A OR 2 x -4.9.
AND WE'LL SIMPLIFY THIS 
IN PIECES.
WE'LL HAVE T = -310 + OR - 
SQUARE ROOT OF THE VALUE
OF THIS EXPRESSION HERE,
WHICH IS THE DISCRIMINATE, 
WE'LL COME BACK TO THAT,
DIVIDED BY -9.8.
SO NOW WE'LL GO 
TO THE CALCULATOR
TO DETERMINE THE VALUE 
OF THE DISCRIMINATE
OR THIS EXPRESSION HERE.
SO WE HAVE 310SQUARED - 4 
x -4.9 x 332.
SO THE DISCRIMINATE
OR THE NUMBER UNDERNEATH 
THE SQUARE ROOT IS 102,607.2.
NOW, NOTICE BECAUSE OF THE 
+ OR - WE ARE GOING TO HAVE
2 SOLUTIONS,
BUT REMEMBER T REPRESENTS 
THE TIME IN SECONDS
AND T CAN'T BE NEGATIVE.
SO ONE SOLUTION 
IS GOING TO BE NEGATIVE,
WHICH DOESN'T MAKE SENSE,
SO WE'LL EXCLUDE THAT ONE
AND THEN WE'LL ONLY USE 
THE POSITIVE VALUE OF T
FOR THE TIME
WHEN THE ROCKET SPLASHES 
INTO THE OCEAN.
SO WE'LL PUT THE NUMERATOR 
IN PARENTHESIS.
SO WE'LL HAVE 
IN PARENTHESIS -310.
LET'S USE + THE SQUARE ROOT 
OF OUR DISCRIMINATE.
1 CLOSED PARENTHESIS 
FOR THE SQUARE ROOT
AND ANOTHER ONE 
FOR THE NUMERATOR
AND THEN WE'LL DIVIDE THIS 
BY -9.8.
NOTICE HOW THIS VALUE 
IS NEGATIVE?
AND SINCE TIME 
CAN'T BE NEGATIVE
WE'RE GOING TO USE 
THE SECOND SOLUTION.
SO WE'RE GOING TO CHANGE 
THIS PLUS SIGN HERE
TO A MINUS SIGN.
WE CAN DO THAT 
BY PRESSING SECOND ENTER,
WHICH BRINGS UP 
THE PREVIOUS EXPRESSION
WHERE WE CAN NOW EDIT IT.
SO I'LL CHANGE THIS PLUS SIGN 
TO A MINUS SIGN
AND THEN PRESS ENTER.
SO THE TIME'S GOING TO BE 
APPROXIMATELY 64.3 SECONDS.
AND THEN 
FOR THE SECOND QUESTION,
TO DETERMINE 
HOW HIGH ABOVE SEA LEVEL
THE ROCKET WILL REACH
WE'LL HAVE TO FIND THE VERTEX 
OF OUR PARABOLA
OR THE X COORDINATE 
OR T COORDINATE
IN THIS CASE IS -B/2A
AND THE Y COORDINATE 
WOULD BE THE FUNCTION VALUE
WHEN T IS = TO -B/2A.
SO THE FIRST STEP,
WE NEED TO DETERMINE THE 
X COORDINATE OF OUR VERTEX,
WHICH IS GOING TO BE 
= TO -B/2A.
SO WE'LL HAVE -310 
DIVIDED BY 2 x -4.9.
WE'LL DO THIS 
ON THE CALCULATOR.
WE DO NEED PARENTHESIS AROUND 
THIS PRODUCT SO 2 x -4.9.
SO WE'LL SAY 
IT'S APPROXIMATELY 31.63.
AGAIN, THIS IS THE TIME IT 
TAKES TO REACH THE HIGH POINT
ABOVE SEA LEVEL,
BUT THE QUESTION IS WHAT 
IS THE FUNCTION VALUE
OR WHAT IS THE HEIGHT 
ABOVE SEA LEVEL?
SO OUR ULTIMATE GOAL 
IS TO FIND H OF 31.63.
SO WE'RE GOING TO SUBSTITUTE 
31.63 FOR T IN OUR FUNCTION
AND THIS WILL GIVE US 
OUR MAXIMUM HEIGHT.
SO WE'LL GO BACK TO 
THE CALCULATOR ONE MORE TIME
AND DETERMINE 
THIS FUNCTION VALUE.
SO THE MAXIMUM HEIGHT 
IS GOING TO BE APPROXIMATELY
5,235.1 METERS 
ABOVE SEA LEVEL.
OKAY, SO I THINK 
WE'RE DONE NOW.
WE'VE ANSWERED BOTH QUESTIONS.
I HOPE YOU FOUND THIS HELPFUL.
