Question

An archer releases an arrow from a shoulder height of 1.39 m. When
the arrow hits the target 18 m away, it hits point A. When the target is
removed, the arrow lands 45 m away. Find the maximum height of the
arrow along its parabolic path.

Answer 1

Ok, we have these points in the parabola: (0, 1.39) and (45, 0)

Answer 2

but we have no information about initial velocity, angle or the height of the target

Answer 3

We need to know at least initial velocity or the height of the target in order to obtain another point in the parabola, and, with three points, we can find the parabola equation and find its maximum height.

Answer 4

i don't see how to solve that, this question does not give any other information or a picture?

Answer 5

http://mastmedical.org/ourpages/auto/2013/12/12/52035199/Chapter%204%20Project.pdf

Answer 6

that's a link a the whole thing. I'm on activity 2.

Answer 7

Ohh now we have a picture!

Answer 8

So do you think you can help me?

Answer 9

sorry, i don't see how to find the height of the target...

Answer 10

how u did the Activity 1?

Answer 11

how u found the parabola in Activity 1?

Answer 12

" hits the target at a point 5 ft above the ground"
@m4thM1nd @emily0824

Answer 13

@emily0824
You still there?

Answer 14

yes, I'm confused tho

Answer 15

Can you explain what's confusing you?

Answer 16

I'm still not sure how to find the maxium height

Answer 17

You need to find the equation of the trajectory before you can find the maximum height.
Do you agree with the 5' height of the target? The information is given in a separate section, but it seems that it is the only information available.

Answer 18

okay!
thanks!

Answer 19

Are you able to find the trajectory?

Answer 20

You have three points, and the standard equation of the trajectory is
\(f(x)=a(x-h)^2+k\)
is this familiar to you?

Answer 21

with three unknowns to be found: a, h and k.

Answer 22

We will assume y=0 at ground level.

Answer 23

yes.. im familiar with that

Answer 24

Using the three known points (0,1.39), (18, 1.524), and (45, 0)
we should be able to solve for the three unknowns a, h and k. ok so far?

Answer 25

yes

Answer 26

Do you know the physical interpretation of a, h and k?

Answer 27

In any case, (h,k) represents the vertex, a is a stretching/compression factor.
You would substitute the three points in the equations and solve for a, h and k.
1.39=(0-h)^2+k
1.524=(18-h)^2+k
0=(45-h)^2+k
The strategy is to eliminate k from the equations and end up with two equations. Then you solve for a and h.
Remember, (h,k) is your vertex.
Good luck!

Answer 28

im not too sure, to be honest i moved schools and this is something i had never done at my old school and this was a chapter i missed. thank you so much for helping!

Answer 29

Try to work on this and if you need further help, just post what you have. We can check your work. But it is important that you do the work so you don't get stuck on the next problem. The more you work in math, the easier math becomes.