- THE FUNCTION H OF T = - 16T SQUARED + 72T + 40
MODELS THE HEIGHT AND FEET OF A BALL
AND T IS TIME IN SECONDS.
WE WANT TO ANSWER THE FOLLOWING QUESTIONS.
NUMBER ONE, HOW HIGH IS THE BALL WHEN THE BALL IS THROWN?
NUMBER TWO, WHAT IS THE MAXIMUM HEIGHT OF THE BALL?
AND THEN NUMBER THREE,
HOW MANY SECONDS IS THE BALL IN THE AIR
BEFORE HITTING THE GROUND.
TO GET A BETTER IDEA OF WHAT'S HAPPENING HERE,
LET'S TAKE A LOOK AT A GRAPH OF THIS FUNCTION.
BUT FIRST RECOGNIZE THAT WE HAVE A QUADRATIC FUNCTION
SO THE GRAPH WILL BE A PARABOLA
AND SINCE "A" IS NEGATIVE, THE PARABOLA WILL OPEN DOWN.
SO HERE'S THE GRAPH OF OUR FUNCTION.
AGAIN THIS SHOULD BE THE T AXIS WHICH IS TIME IN SECONDS
AND THE Y AXIS IS HEIGHT AND FEET.
SO WE CAN ALMOST ANSWER ALL THESE QUESTIONS
FROM THE GRAPH.
BUT WE CAN SEE THAT THE BALL IS THROWN AT THIS INSTANT
WHEN THE TIME IS EQUAL TO ZERO.
WE CAN SEE THE INITIAL HEIGHT WOULD BE 40,
THEN IT GOES UP WHICH IS THE MAXIMUM HEIGHT HERE
WHICH SHOULD BE THE VERTEX OF THE PARABOLA.
REMEMBER THE COORDINATES OF THE VERTEX ARE - B/2A COMMA
IN THIS CASE IT WOULD BE H OF - B/2A.
AND THEN HERE IS WHERE IT HITS THE GROUND
AND WE CAN SEE IT'S GOING TO HIT THE GROUND
AT FIVE SECONDS.
WELL LET'S TAKE A LOOK AT THE CALCULATIONS
TO DETERMINE THESE VALUES.
FIRST TO DETERMINE HOW HIGH THE BALL IS
WHEN THE BALL IS THROWN,
AT THE INSTANT THE BALL IS THROWN,
IS WHEN THE TIME STARTS.
SO IF WE SET T TO ZERO WE CAN FIND THE INITIAL HEIGHT.
SO IF T IS EQUAL TO ZERO,
H OF ZERO WOULD BE THE INITIAL HEIGHT
WHICH WOULD BE - 16 x 0 SQUARED + 72 x 0 + 40
WHICH WE CAN SEE WOULD JUST BE 0 + 0 + 40 OR 40 FEET
WHICH AGAIN WE SAW FROM THE GRAPH
THE INITIAL HEIGHT WAS 40 FEET.
THE SECOND QUESTION IS TO DETERMINE
THE MAXIMUM HEIGHT OF THE BALL
WHICH REMEMBER WOULD OCCUR AT THE VERTEX
WHERE THE X COORDINATE OR THE T COORDINATE
WOULD BE THE TIME IT TAKES TO REACH THE MAXIMUM HEIGHT
AND THE Y COORDINATE WOULD ACTUALLY BE THE HEIGHT
OF THE BALL.
SO TO DETERMINE THE Y COORDINATE OF THE VERTEX
WE FIRST HAVE TO DETERMINE THE X COORDINATE
OR IN THIS CASE THE T COORDINATE
AND THEN WE'LL EVALUATE THE FUNCTION AT THAT VALUE OF T.
SO IN LOOKING AT OUR FUNCTION,
"A": IS EQUAL TO - 16 AND B IS EQUAL TO 72.
SO IF THE X COORDINATE OR T COORDINATE
WOULD BE EQUAL TO - 72 DIVIDED BY 2 x - 16
WHICH WOULD BE NEGATIVE 72 DIVIDED BY - 32
WHICH COMES OUT TO EXACTLY 2.25
WHICH AGAIN REMEMBER IS TIMES, THIS IS SECONDS.
THIS IS HOW LONG IT TAKES FOR THE BALL
TO REACH THE MAXIMUM HEIGHT.
BUT THE QUESTION ASKS FOR THE MAXIMUM HEIGHT.
NOW WE NEED TO FIND H OF 2.25.
SO WE'LL HAVE - 16 x 2.25 SQUARED + 72 x 2.25 + 40
AND NOW WE'LL USE THE CALCULATOR.
SO WE'LL HAVE - 16 x 2.25 SQUARED + 72 x 2.25 + 40.
SO THE MAXIMUM HEIGHT IS 121 FEET
WHICH IS THE ANSWER WE'RE LOOKING FOR.
OKAY, AND THEN FOR THE THIRD QUESTION,
WE WANT TO KNOW HOW MANY SECONDS THE BALL IS IN THE AIR
BEFORE HITTING THE GROUND.
AGAIN GOING BACK TO THE GRAPH WE CAN SEE IT'S EQUAL TO 5,
BUT LET'S TAKE A LOOK AT THE ALGEBRA BEHIND THIS.
WHEN THE BALL HITS THE GROUND, THE HEIGHT WOULD BE 0
OR HFT WOULD BE EQUAL TO 0.
SO TO ANSWER THIS QUESTION ALGEBRAICALLY
WE'D HAVE TO SOLVE THE EQUATION
0 = - 16T SQUARED + 72T + 40.
LET'S SEE IF WE CAN FACTOR THIS.
WE'LL FIRST FACTOR OUT THE GREATEST COMMON FACTOR OF 8.
BUT IF WE WANT THE COEFFICIENT TO BE POSITIVE,
LET'S FACTOR OUT - 8.
THAT WOULD LEAVE US WITH POSITIVE 2T SQUARED - 9T
AND THEN -5.
REMEMBER IF WE FACTOR OUT A NEGATIVE
IT'S GOING TO CHANGE THE SIGN OF EACH TERM.
SO IF THIS DOES FACTOR,
WE'LL FACTOR INTO TWO BINOMIAL FACTORS
THE FIRST TERMS MUST COME FROM THE FACTORS OF 2T SQUARED,
THAT WOULD BE 2T AND T.
THE SECOND TERMS MUST COME FROM THE FACTORS OF - 5
SO THAT THE SUM OF THE INNER PRODUCT AND OUTER PRODUCT
WOULD BE - 9T.
NOTICE IF WE PUT MINUS 5 HERE,
WE'D HAVE TO PUT A POSITIVE 1 HERE.
THE OUTER PRODUCT IS - 10T,
THE INNER PRODUCT IS POSITIVE 1T,
WHICH DOES HAVE A SUM OF - 9T.
SO THIS IS FACTORED CORRECTLY.
SOLUTIONS WOULD BE WHEN 2T + 1 = 0 OR 1T - 5 = 0.
HERE WOULD SUBTRACT 1 AND THEN DIVIDE BY 2,
THAT WOULD BE T = - ONE HALF.
HERE WE'D ADD 5 SO WE HAVE T = 5.
REMEMBER T IS TIME IN SECONDS.
TIME CAN'T BE NEGATIVE SO THIS DOES NOT APPLY.
OUR ONLY SOLUTION IS T = 5 OR 5 SECONDS.
OKAY. HOPE YOU FOUND THIS HELPFUL.