A ball is thrown off a two-story house
24 feet above the ground.
Its starting velocity, also
known as the initial velocity,
is negative six feet per second.
The negative means its initial velocity
is downward toward the ground.
The equation H equals negative
16T squared minus 6T plus 24
can be used to model the height H
of the ball after T seconds.
About how long does it take
for the ball to hit the ground?
When the ball is on the ground,
the height H is equal to zero,
and therefore, to answer this question,
we substitute zero for H and solve for T.
Performing this substitution,
we have the equation
zero equals negative 16T
squared minus 6T plus 24.
Let's solve this quadratic equation using
the quadratic formula shown here,
where in this form, A equals negative 16,
B equals negative six, and C equals 24.
But instead of using these
values for A, B and C,
let's go ahead and factor
the greatest common factor
from the right side of the equation.
The greatest common factor would be two,
but instead of factoring out two,
let's factor out negative two so the
co-efficient of the T squared
term will be positive.
So if we factor negative
two from the right side,
our equation is zero
equals negative two times.
If we factor out negative two
from negative 16 T squared,
we'll have a positive 8T squared.
Notice if we multiply, we still
have negative 16T squared.
We factor negative two from negative 6T.
We'll end up with positive 3T.
So we have plus 3T.
If we factor negative
two from positive 24,
we're left with negative
12, and therefore,
we have minus 12.
Notice how this equation is equal to zero
when the quadratic into the
parentheses is equal to zero.
So now we can use the quadratic formula
and use A equals eight, B equals three,
and C equals negative 12.
However, if we wanted to, we could also
divide both sides of the
equation by negative two.
If we simplify, zero divided by
negative two is still zero.
So if zero equals, on
the right, negative two
divided by negative two simplifies to one.
And therefore, we have zero equals
8T squared plus 3T minus 12.
Let's go ahead and solve
the equation in this form.
When in this form, A
equals positive eight,
B equals positive three
and C equals negative 12.
Let's continue on the next slide.
The next step is to substitute A, B and C
into the quadratic formula.
But of course, instead
of X on the left side,
we will have the variable T.
So we have T equals...
in the numerator, negative B
is going to be negative three,
and we have plus or minus the square root
of B squared is the square of three
minus four times A times C is minus four
times eight times negative 12.
All this is divided by two times A,
which is equal to two times eight.
And now let's begin simplifying.
We have T equals, in the numerator,
we have negative three plus or minus
the square root.
We need to be careful when
simplifying the discriminant.
We simplify the exponents first.
Three squared is equal to nine.
And then we have minus four
times eight times negative 12.
Four times eight times negative 12
equals negative 384, but we
still have this minus here.
Minus 384 is equivalent to plus 384.
The denominator is two times
eight, which equals to 16.
Nine plus 384 is equal to 393.
So we have t equals
negative three plus or minus
the square root of 393, all divided by 16.
The square root of 393 does not simplify.
So one solution is T
equals negative three plus
the square root of 393 divided by 16.
Another solution is T
equals negative three minus
the square root of 393 divided by 16.
Let's break this up into
two separate fractions.
So here we have negative
3/16 plus the square root
of 393 divided by 16, and here we have
negative 3/16 minus the square
root of 393 divided by 16.
Now we have to get our
decimal approximations
for both of these using a calculator.
So for the first value of
T, we enter negative three
divided by 16, and then
plus the square root,
which in this calculator,
is second X squared.
And then we have 393.
Right arrow to get
outside the square root,
and again, divide by 16.
Enter.
To two decimal places.
T is approximately 1.05.
And this would be seconds.
And now the only difference for the second
value of T is instead
of plus, we have minus.
So on this calculator, we
can press second enter,
which brings up the previous entry,
and an arrow to the left and
change the plus to a minus.
And enter.
To two decimal places.
We have T is approximately negative 1.43.
So both of these values
of T are approximate
solutions to this algebraic equation
or this quadratic equation.
But remember, T represents time
and time cannot be negative,
and therefore, the ball
will hit the ground
in approximately 1.05 seconds.
Let's write the answer
as a complete sentence.
I hope you found this helpful.
