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## Algebra 1 (Eureka Math/EngageNY)

### Unit 4: Lesson 12

Topic B: Lesson 17: Graphing quadratic functions from the standard form- Finding the vertex of a parabola in standard form
- Graphing quadratics: standard form
- Graph quadratics in standard form
- Quadratic word problem: ball
- Quadratic word problems (standard form)
- Forms & features of quadratic functions
- Worked examples: Forms & features of quadratic functions
- Features of quadratic functions: strategy
- Finding features of quadratic functions
- Warmup: Features of quadratic functions
- Features of quadratic functions
- Graph parabolas in all forms

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# Quadratic word problem: ball

CCSS.Math: , ,

Sal solves a word problem about a ball being shot in the air. The equation for the height of the ball as a function of time is quadratic. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What if you're trying to find the maximum height and not how long it's in the air?(48 votes)
- You'd want to find the vertex of the parabola. See the graph I made in the question above for an approximate answer.(42 votes)

- where did the 16t^2 come from?(19 votes)
- Are you sure that -16^2 is -32? I think it is 256.(1 vote)

- What would you do if you were using the quadratic formula and came up with a negative square root?(12 votes)
- Then you would say that there are no real solutions to the quadratic equation. But when you get into higher levels of math, you will come across imaginary numbers, and there will be a solution. But for now, just say that there aren't any real solutions.(19 votes)

- How would this problem be solved using the "completing the square" method instead of the quadratic formula?(16 votes)
- Actually, the quadratic formula is derived from the completing the square method and so they are essentially, the same. They always work. quadratic formula derivation video

https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/v/proof-of-quadratic-formula(8 votes)

- We could take h to be 0 or -50. Dependent on perspective. Since there are 2 different perspectives, how do we determine whether it's 0 or -50?(5 votes)
- Often the simplest perspective is the one to use, and there may be clues in the words that are used. A good technique is to try to sketch the circumstances in the problem and then think carefully about what's happening.

The problem says you are 50 feet ABOVE the ground. So in a drawing, that point would normally be +50.

The problem also asks how long will it take for the ball to HIT THE GROUND. So, after "t" seconds, it has come back down to earth with no "height" remaining. That's why Sal used zero in the equation.

I hope this is helpful! (I've never found word problems easy!)(10 votes)

- Does be gravitation acceleration in this exercise 32m/s^2 ?

I know its a exercise about quadratic equation not kinematic, but i it can confuse.

Just look:

s = Vavg. * Δt = (Vi + Vf)/2 * Δt = (Vi + Vi + a*Δt)/2 * Δ = Vi*Δt + (a*Δt^2)/2

then we will subtitute the variables with values

Vi = 20m/s, a = -32m/s^2 (beacuse its accelerating to the ground), s= -50m (negative because we need 50 meters to reach the ground)

-50m = 20m/s * Δt + (-32m/s^2 * Δt^2)/2

-50m = 20m/s * Δt - 16m/s^2 * Δt^2

0 = 20m/s * Δt - 16m/s^2 * Δt^2 + 50m // if we leave the Δ symbol, and m/s we will get...

0 = -16t^2 + 20t + 50(5 votes)- On Earth, acceleration due to gravity is 9.8 m/s². This corresponds to 32.2 ft/s².(5 votes)

- Can someone please help me understand this problem?:

A ball is thrown directly upward from a height of 30 feet with an initial velocity of 64 feet per second. The equation h=-16t^2+64t+30 gives the height h after t seconds.

Solve for both t and h.(5 votes)- Maximum height h -> find the axis of symmetry -> x= -b/2a= -64/2(-16)= 2, plug it in -> -16(2)^2+64(2)+30= 94, so maximum height is 94 ft.

At what time will the ball reach the ground? Set h equal to 0 -> 0=-16t^2+64t+30, solve for t,*+- means plus or minus*-> (-64+-sqrt(64^2-4*-16*30))/2*-16, t=-0.424s and t=4.424s. Since time**cannot**be negative, it will take the ball 4.424 seconds to hit the ground.(4 votes)

- What if I would want to know at what time would the ball pass certain point?

For example, if I would want to know at what time would the ball pass the same height--going down-- as the initial launch?

Would an equation like 50 = - 16t^2 - 20t + 50 give me an answer?(4 votes)- Yes, you can do that. The equation has a variable for height (h), so you could plus in any value that you want.(4 votes)

- Can the problem be solved by using completing the square? Why?(4 votes)
- Yes you can. Completing the square works on any quadratic equation. It can get take a lot of time to do it but it always works.(2 votes)

- I am working on Solving Applications Involving Distance, Rate, and Time, (in a word problem) I am having a very hard time for some reason grasping how to turn time into a fraction. ex: 5 hours and 20 mins. I want to write it as 320/60 but I know that is not right.(2 votes)
- That's correct if the unit is hours, but you need to figure out what unit you want.

5 h + 20 min = 320 min = 320 / 60 h = 16 / 3 h(4 votes)

## 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.