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## Quadratic standard form

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# Quadratic word problem: ball

CCSS.Math: , ,

## Video transcript

A ball is shot into the air
from the edge of a building, 50 feet above the ground. Its initial velocity
is 20 feet per second. The equation h--
and I'm guessing h is for height-- is equal to
negative 16t squared plus 20t plus 50 can be used to
model the height of the ball after t seconds. And I think in this
problem they just want us to accept this
formula, although we do derive formulas
like this and show why it works for this type of
problem in the Khan Academy physics playlist. But for here, we'll just go
with the flow on this example. So they give us
the equation that can be used to model the height
of the ball after t seconds, and then say about
how long does it take for the ball
to hit the ground. So if this is the
height, the ground is when the height
is equal to 0. So hitting the ground
means-- this literally means that h is equal to 0. So we need to figure out at
which times does h equal 0. So we're really
solving the equation 0 is equal to negative 16t
squared plus 20t plus 50. And if you want to simplify
this a little bit-- let's see, everything here is
divisible at least by 2. And let's divide
everything by negative 2, just so that we can get rid
of this negative leading coefficient. So you divide the left
hand side by negative 2, you still get a 0. Negative 16 divided
by negative 2 is 8. So 8t squared. 20 divided by negative
2 is negative 10. Minus 10t. 50 divided by negative
2 is minus 25. And so we have 8t squared minus
10t minus 25 is equal to 0. Or if you're comfortable with
this on the left hand side, we can put on the
left hand side. We could just say
this is equal to 0. And now we solve. And we could complete
this square here, or we can just apply the
quadratic formula, which is derived from
completing the square. And we have this
in standard form. We know that this is our a. This right over here is our b. And this over here is our c. And the quadratic formula
tells us that the roots-- and in this case, it's in
terms of the variable t-- are going to be
equal to negative b plus or minus the
square root of b squared minus 4ac, all of that over 2a. So if we apply it, we get
t is equal to negative b. b is negative 10. So negative negative 10 is
going to be positive 10. Plus or minus the square
root of negative 10 squared. Well, that's just positive
100, minus 4 times a, which is 8, times c,
which is negative 25. And all of that over 2a. a Is 8. So 2 times 8 is 16. And this over here,
we have a-- let's see if we can simplify
this a little bit. The negative sign,
negative times a negative, these are going to be positive. 4 times 25 is 100,
times 8 is 800. So all that simplifies to 800. And we have 100 plus 800
under the radical sign. So this is equal
to 10 plus or minus the square root of 900,
all of that over 16. And this is equal to 10 plus
or minus-- square root of 900 is 30-- over 16. And so we get time is equal
to 10 plus 30 over 16, is 40 over 16, which is
the same thing if we divide the numerator and denominator
by 4 to simplify it as 10 over-- or actually even better, divide
it by 8-- that's 5 over 2. So that's one solution,
if we add the 30. If we subtract the 30,
we'd get 10 minus 30. Or t is equal to 10 minus 30,
which is negative 20 over 16. Divide the numerator and
the denominator by 4, you get negative 5 over 4. Now, we have to remember,
we're trying to find a time. And so a time, at
least in this problem that we're dealing
with, we should only think about positive times. We want to figure out
how much time has taken-- how long does it take for
the ball to hit the ground? We don't want to
go back in time. So we don't want our
negative answer right here. So we only want to think
about our positive answer. And so this tells us that the
only root that should work is 5/2. And we assume that
this is in seconds. So this is in 5/2 seconds. I wouldn't worry too much
about the physics here. I think they really just want us
to apply the quadratic formula to this modeling situation. The physics, we go
into a lot more depth and give you the conceptual
understanding on our physics playlist. But let's verify that we
definitely are at a height of 0 at 5/2 seconds, or
t is equal to 5/2. This expression right over here
does give us h is equal to 0. So we have-- let's try it out. We have negative 16
times 5/2 squared plus 20 times 5/2 plus 50. This needs to be equal to 0. So this is negative
16 times 25/4 plus-- let's see, if we divide
20 by 2, we get 10. If we divide 2 by 2, we get 1. So 10 times 5 is going to be 50. Plus 50. This needs to be equal to 0. Negative 16 divided
by 4 is negative 4. 4 divided by 4 is 1. So you have negative 4
times 25, which is 100. Plus 50-- oh, sorry. Negative 4 times
25 is negative 100. Plus 50, plus 50
again is equal to 0. And so we have negative
100 plus another 100. Well, that's definitely
going to be equal to 0. We get 0 equals 0. And it all checks out. We hit the ground
after 5/2 seconds. Or another way to think
about it is 2.5 seconds. t is equal to 5/2
seconds, or 2.5 seconds.