College algebra, several questions about height, speed, etc.

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Boots

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Hi, this is my first post here, please be gentle. :) I read the rules, and I know it says don't post a big list, but it also says post the entire problem, and my entire problem is a big list, so I'm sorry about that. I'll put the problem and questions, (it's all part of the same project) and the stuff in parenthises is what I've been able to guess at, but I really have no idea where to even start, or what to use, or anything.

Using the quadratic function: h(t)=at[sup:qc9too2i]2[/sup:qc9too2i]+v[sub:qc9too2i]0[/sub:qc9too2i]+h[sub:qc9too2i]0[/sub:qc9too2i]

An archer on a hill 17ft above the ground shoots an arrow at a rate of 280ft/sec. (No angle was listed anywhere, so I'm guessing it's just horizontal?)

a) What is the initial height? (17ft?)
b) What is the initial velocity? (Zero because it wasn't moving at first?)
c) What is the equation modeling the path? (No idea here. Is this where you take that above equation and plug things in?)
d) Draw graph
e) What is the height of the arrow above the ground 3 seconds after the arrow was shot?
f) After how many seconds does the arrow reach max height? (If there's no angle and it was shot horizontally, wouldn't the starting height be the max? It can only go down from there...)
g) What is max height? (Ditto^)
h) How long did it take for the arrow to hit the ground?
i) Two possible answers, why is one wrong? (One will be negative, no negatives allowed)
 
Re: College algebra, several questions about height, speed,

Boots said:
Hi, this is my first post here, please be gentle. :) I read the rules, and I know it says don't post a big list, but it also says post the entire problem, and my entire problem is a big list, so I'm sorry about that. I'll put the problem and questions, (it's all part of the same project) and the stuff in parenthises is what I've been able to guess at, but I really have no idea where to even start, or what to use, or anything.

Using the quadratic function: h(t)=at[sup:2iqbhi2m]2[/sup:2iqbhi2m]+v[sub:2iqbhi2m]0[/sub:2iqbhi2m]+h[sub:2iqbhi2m]0[/sub:2iqbhi2m]

An archer on a hill 17ft above the ground shoots an arrow at a rate of 280ft/sec. (No angle was listed anywhere, so I'm guessing it's just horizontal?)

a) What is the initial height? (17ft?) ......................................................Correct
b) What is the initial velocity? (Zero because it wasn't moving at first?) .....Yes ......... the initial velocity is horizontal velocity
c) What is the equation modeling the path? (No idea here. Is this where you take that above equation and plug things in?) .............. Yes
d) Draw graph
e) What is the height of the arrow above the ground 3 seconds after the arrow was shot? ................. evaluate 'h' at t= 3
f) After how many seconds does the arrow reach max height? (If there's no angle and it was shot horizontally, wouldn't the starting height be the max? It can only go down from there...) .............. Yes

g) What is max height? (Ditto^) .............. Yes

h) How long did it take for the arrow to hit the ground? .............. find values of 't' for h = 0 ? quadratic equation
i) Two possible answers, why is one wrong? (One will be negative, no negatives allowed)
Click to expand...

Tell us where you are stuck

Since your problem did not specify - it is poorly worded problem. They could have meant that the arrow was shot straight-up - vertically.
 
Re: College algebra, several questions about height, speed,

Sorry... I'm having trouble here:

c) equation modeling path: Would h(t)=at[sup:2lfmirto]2[/sup:2lfmirto]+v[sub:2lfmirto]0[/sub:2lfmirto]+h[sub:2lfmirto]0[/sub:2lfmirto] turn into 17(280)=-16(280)[sup:2lfmirto]2[/sup:2lfmirto]+v[sub:2lfmirto]0[/sub:2lfmirto]+17[sub:2lfmirto]0[/sub:2lfmirto]? If so, I'm not sure what to do with the zeros next to v and the last 17.

d) Will the above equation gives points for a graph? Otherwise I'm lost on how to graph it. I have a general idea of what it should look like, but I don't know how to get the points.

e) It would turn into h(3)=(-16)3[sup:2lfmirto]2[/sup:2lfmirto]+v[sub:2lfmirto]0[/sub:2lfmirto]+h[sub:2lfmirto]0[/sub:2lfmirto], right? Maybe? Again, not sure what to do with the subscripts...

h) Solve for t when h=0... So 0=at[sup:2lfmirto]2[/sup:2lfmirto]+v[sub:2lfmirto]0[/sub:2lfmirto]+0[sub:2lfmirto]0[/sub:2lfmirto] is as far as I get.

Also, I'm putting -16 for a because that's what the paper said a equalled... -16 ft/sec[sup:2lfmirto]2[/sup:2lfmirto] (But instead of putting -16 should I be putting -16[sup:2lfmirto]2[/sup:2lfmirto]?)
 
Boots said:
Using the quadratic function: h(t) = at[sup:1kib4ooc]2[/sup:1kib4ooc] + v[sub:1kib4ooc]0[/sub:1kib4ooc] + h[sub:1kib4ooc]0[/sub:1kib4ooc] ? This is not correct

The middle term is missing a factor of t.

h(t) = a*t^2 + v_0*t + h_0
Click to expand...
 
Boots said:
Would h(t)=at[sup:1ic4q8lf]2[/sup:1ic4q8lf]+v[sub:1ic4q8lf]0[/sub:1ic4q8lf]+h[sub:1ic4q8lf]0[/sub:1ic4q8lf] turn into 17(280)=-16(280)[sup:1ic4q8lf]2[/sup:1ic4q8lf]+v[sub:1ic4q8lf]0[/sub:1ic4q8lf]+17[sub:1ic4q8lf]0[/sub:1ic4q8lf]?

NO

The symbolism h(t) is function notation for y.

In other words, the entire symbol h(t) is a variable; it represents a number.

h(t) does not mean h times t. The letter h is only the name of the function, here.

The parentheses show the input.

EG:

h(2) is function notation for the output when t = 2

h(5) is function notation for y when t = 5

h(9) represents the height after 9 seconds

et cetera


If so, I'm not sure what to do with the zeros next to v and the last 17.

We do not do any arithmetic with those zeros; they're subscripts.

I type v[sub:1ic4q8lf]0[/sub:1ic4q8lf] like this: v_0. Same with h_0.

Generally, when a variable (or constant) symbol has zero as a subscript, it represents the initial value.

In this exercise, time starts out at t = 0; therefore, v_0 is the velocity at time zero, and h_0 is the height of the arrow at time zero.

With the assumption that the arrow is shot horizontally, there is no vertical velocity. Therefore, replace the symbol v_0 with zero, and totally ignore the number 280.

Replace the symbol h_0 with the initial height.


d) Will the above equation gives points for a graph?

Yes. You pick values for t (the function input) and use the height formula to calculate the corresponding values of h(t).

Those ordered pairs of numbers are the coordinates of points on the parabola.


e) It would turn into h(3)=(-16)3[sup:1ic4q8lf]2[/sup:1ic4q8lf]+v[sub:1ic4q8lf]0[/sub:1ic4q8lf]+h[sub:1ic4q8lf]0[/sub:1ic4q8lf], right?

Sorta. Substitute:

t = 3

v_0 = 0

h_0 = 17

Then, do the math, to determine the value of the constant h(3).


I'm putting -16 for a because that's what the paper said

But instead of putting -16 should I be putting -16[sup:1ic4q8lf]2[/sup:1ic4q8lf]?

NO

a = -16, not (-16)^2

When you see a term like at^2, it means that the exponent applies only to the value of t.

Remember the Order of Operations ?

If they wanted both of the numbers a and t to be squared, they would be required to use grouping symbols to show this.

(at)^2
Click to expand...

Thank you so much for showing your work or explaining what you're thinking.

Most people seeking help here seem to expect us to read their mind.

Cheers ~ Mark 8-)

 


I suggest that you ask your instructor for clarification on the direction of the arrow.

If the arrow is shot vertically, then v_0 = 280.

I'm thinking that the arrow is shot vertically; otherwise, the arrow hits the ground after only 1.03 seconds of flight.

In other words, with the assumption that the arrow is shot horizontally, the value of h(3) is negative. I don't think we can have negative height, in this exercise.

 
Re: College algebra, several questions about height, speed,

I definitely agree with the previous responder...a much more NORMAL application of the given formula would be when the projectile is shot upward.

The initial height h[sub:dpeumjua]0[/sub:dpeumjua] would be the height above level ground from which the arrow is shot (17 ft).

The initial velocity v[sub:dpeumjua]0[/sub:dpeumjua] is 280 ft/s

Then, you'd normally be looking for when the arrow hits the ground (the value of "t" when h(t) = 0) or what is the maximum height the arrow reaches.

Please check with your teacher...it isn't wise for us to ASSUME we know what your problem REALLY asks for.
 
Re: College algebra, several questions about height, speed,

\(\displaystyle Assuming \ arrow \ is \ shot \ straight \ up, \ then:\)

\(\displaystyle S(t) \ = \ -16t^2+V_0(t)+S_0 \ and \ S'(t) \ = \ V(t) \ = \ -32t+V_0.\)

\(\displaystyle V_0 \ = \ initial \ velocity \ = \ 280ft./sec. \ and \ S_0 \ = \ initial \ height \ = \ 17 \ ft.\)

\(\displaystyle a) \ S(0) \ = \ -16(0^2)+280(0)+17 = 17 ft.\)

\(\displaystyle b) \ V(0) \ = \ -32(0)+280 \ = \ 280 \ ft./sec.\)

\(\displaystyle c) \ S(t) \ = \ -16t^2+280t+17\)

\(\displaystyle d) \][attachment=1:yoniebj6]ccc.jpg[/attachment:yoniebj6]\)

\(\displaystyle e) \ S(3) \ = \ -16(3^2)+280(3)+17 \ = \ 713 \ ft.\)

\(\displaystyle f) \ V(t) \ = \ 0 \ = \ -32t+280, \ t \ = \ 8.75 \ sec.\)

\(\displaystyle g) \ S(8.75) \ = \ -16(8.75^2)+280(8.75)+17, \ h \ = \ 1,242 \ ft.\)

\(\displaystyle h) \ S(t) \ = \ 0 \ = \ -16t^2+280t+17, \ t \ \dot= \ 17.56 \ seconds.\)

\(\displaystyle i) \ No \ negatives \ allowed.\)

\(\displaystyle Note, \ in \ actuality, \ (neglecting \ wind \ resistance) \ the \ arrow \ would \ come \ down \ and \ stick \ in\)

\(\displaystyle your \ head; \ however \ we \ expanded \ the \ above \ graph \ to \ account \ for \ time, \ the \ more \ realistic\)

\(\displaystyle graph \ is \ shown \ below.\)

[attachment=0:yoniebj6]ddd.jpg[/attachment:yoniebj6]

\(\displaystyle I \ shot \ an \ arrow \ in \ the \ air; \ where \ it \ landed \ I \ know \ not \ where; \ however, \ if \ you \ shoot \ the\)

\(\displaystyle arrow \ straight \ up, \ it \ might \ land \ in \ your \ head, \ so \ be \ careful.\)
 
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