Question

A ball is thrown upward and outward from the top edged of a 50 ft building it reaches its highest point 20 ft above and 10 ft out from the building. How far from the building is the ball when it hits the ground ?

Answer 1

Are you familiar with the motion formulas?

Answer 2

its a physics question...

Answer 3

We are instructed to do it in analytic geometry

Answer 4

mathematical modelling

Answer 5

so it is an application of a parabola

Answer 6

Can I just see your imagination in this problem. I can't imagine it Im noob at applications

Answer 7

I'm solving in the view of physics. Though now you said it, it's probably more wise to use parabola to solve it.

Answer 8

So, you have a parabola, with vertex(10,20) and a point on the parabola, (0,50)

Answer 9

It is stated in the book that the answer must be 28.7 feet

Answer 10

Should it be (60,70) ?

Answer 11

Yeah, it should.

Answer 12

I meant (10 , 70)

Answer 13

Yeah...so the vertex is (10,70) there's a point(0,50), and we're trying to find the x-intercept. Sub all these into the parabola, \((x+k)^2 = 4a(y+h)\)
What did you get?

Answer 14

should it be ? (X - H)^2 = 4a(y-k)

Answer 15

oh. Your convention is negative.

Answer 16

I meant - 4a

because its downward right ?

Answer 17

But this is what confuses me how do you that that's a vertex :(

Answer 18

Yup.Though that will be calc'ed later.
because it's the highest point of the whole journey.
the parabola is \((x-10)^2=-5(y-70)\)
Solving with y=0 gives 28.7.

Answer 19

so 4a = - 5 my god thank you Shadowys

Answer 20

That was done by subbing (0,50) into it.
You're welcome :)

Answer 21

wait so it gives a quadratic equation ?

Answer 22

Yup, it does.

Answer 23

x^2 - 20x - 450?

Answer 24

The answer is not exact -_-

Answer 25

Um, how did you get that equation? from \((x−10)^2=−5(y−70)\)?

Answer 26

That is the parabola of the thrown ball.

Answer 27

X^2 - 20x +100 = -5( 0 - 70)

Answer 28

X^2 - 20x - 250

Answer 29

it's 28.708289 to be exact. Yes, that gives the correct answer.
Though you could save some steps by not expanding.

Answer 30

Aw how did you do it :O

Answer 31

let me guess square root ?

Answer 32

Now I get I FORGOT THE BASICS HAHAHAHAH I forgot to square both sides . Thanks again Shadowy! :)

Answer 33

A ball is thrown upward and outward from the top edged of a 50 ft building it reaches its highest point 20 ft above and 10 ft out from the building.

hmm vertex is as stated (10,50+20); vertex form of the equation is then

y = a(x-10)^2 + 70 ; when x=0, then y=50

50 = a(-10)^2 + 70
-20 = a 100
-.2 = a

therefore the equation would be:

y = -.2(x-10)^2 + 70, so when does y=0?

0 = -.2(x-10)^2 + 70
-70 = -.2(x-10)^2
350 = (x-10)^2
sqrt(350) = x-10
10 + sqrt(350) = x = 28.708...

yep

Answer 34

Lol glad you've got it :)

Answer 35

;) just keeping the old grey matter nice and pliable, good job