- [Phillip] So we're given
that the height of a ball
in meters, t seconds after
it's thrown off a building
is given by negative 9.8 t
squared plus 20 t plus 50.
Very first question is
how tall is the building?
At the instant before the
ball is thrown, so h of zero,
we find out is 50.
So the building is 50 meters tall.
Do not forget your units.
When does the ball hit the ground?
Well that's gonna be when
this is equal to zero,
negative 9.8 t squared plus 20 t plus 50.
This is going to probably
not factor very easily,
so we'll go ahead and use
the quadratic formula.
And I'm gonna go ahead and
I like my coefficient on
t squared to be positive
so 9.8 t squared.
Doesn't need to be, just a habit.
Okay, and we're gonna plug these
into our quadratic formula.
So now I have 20 plus
or minus the square root
of 20 squared minus four
times 9.8 times negative 50.
Remember to bring your signs with those,
all over two times 9.8.
Okay, let's use our calculator
to help us calculate those.
And I'm gonna go ahead just
put everything in at once.
And we have our numerator's 20 plus,
I'm gonna do the plus first,
square root of 20 squared
minus four times 9.8
times negative 50.
There's our numerator divided by,
now because I have the denominator written
I do need to put it in
a set of parenthesis,
two times 9.8.
And I see that it lands,
one of my values is
t is about 3.5 seconds,
it's going to land.
Now we have another value coming from this
and that is when I do 20 minus.
And I do get a value.
I get another value, I get negative 1.46,
but negative doesn't
make sense in the context
of this situation, so
we're not gonna worry about
that answer in this case.
Okay next thing, what is the max height
and when does it reach it.
Well our graph is a quadratic,
so it's maximum height is
gonna be at it's vertex.
To calculate the vertex,
we can use the formula
minus v over two A.
So minus, now V is 20, A is negative 9.8
and that has to get multiplied by two.
So calculating that,
20 divided by, I'm just
gonna put 'em in as positive,
times 9.8.
So the maximum height is going to occur
at roughly about 1.02 seconds.
Now to get the maximum height,
I'm gonna plug that
value into my quadratic.
So this is when it reaches
it's maximum height.
It's maximum height is
gonna come from plugging
1.02 seconds into my function.
So one thing I can do
to save all my decimals,
is right here, down at the
bottom I've got the answer key.
So what I'm gonna do is I'm gonna put in
negative 9.8 times and I'm gonna do second
that answer key at the bottom
that I showed you squared
plus 20 times second answer plus 50.
So that automatically puts
that number in for me.
And there's my maximum
height, about 60.2 meters.
So the max height is about 60.2 meters
above the ground.
Now what about when the ball
is 20 meters above the ground?
Well, remember what our function does.
It gives the height above the ground
as a function of t,
so we're gonna set that equal to 20.
So we're gonna have 20
equals negative 9.8 t squared
plus 20 t plus 50.
So once again I have a quadratic
I'm gonna get equal to zero
and I'm gonna do it by
bringing everything over here.
So 9.8 t squared minus 20
t minus 30 equals zero.
So once again I'm gonna
use my quadratic formula.
So t equals 20 plus or
minus the square root
of 20 squared minus four
times 9.8 times negative 30,
it's very similar to the one we did
just a little bit ago,
over two times 9.8.
In fact, so similar that
I'm gonna use that one.
If I do second entry,
it brings up some of my old answers.
So I do second entry a few times
and there's back to my quadratic formula.
Biggest thing is this guy
that has to get changed
to a negative 30.
And notice this is my answer for minus.
So I do get a negative value for t.
So one of my values is negative 1.005
to a couple of decimal places.
But that doesn't make
sense in this context.
So there's only gonna be one place
where it's gonna be
20 meters that makes
sense in this context.
And let's change that to a plus.
And that's at about 3.04 seconds.
Not too long, before it's
about to hit the ground.
