intermediate algebra

hopen2impov

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Aug 4, 2010
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solve using the five step problem solving process, translate, reword, carry-out, check, and state.
An object altitude, in meters is given by the polynomial ht+vt-9.8t^2 where h is the height in meters from which the launch occurs. v is the initial upward speed in meters per second and t is the number of seconds for which the rocket is airborne. A pebble is shot upward from the top of a building 164 meters tall. If the initial speed is 36 meters per second. How high above ground will the peble be after 3 seconds. (round to the nearest tenth of a meter).

ht +vt -9.8t^2

h=164 meters + vt=36(3) -9.8t^2
h=164 meters + 108 -9.8t^2
h=164 meters +172 meters -9.8t^2
answer 163.8t^2 i do not know this but i hope i'm right
 
hopen2impov said:
ht + vt - 9.8t^2, where h is the height in meters from which the launch occurs.

Ah, this exercise has returned.

I see that you found the missing plus sign between ht and vt.

I see that the coefficient on t^2 (it is -9.8) now matches gravitation near planet Earth.

Yet, I regret to inform you that the given expression is STILL WRONG.

The first term is not ht.

However, this posted version makes enough sense that I'm willing to look at your work. Maybe "ht" is a typographical error.

Hold on.

 
hopen2impov said:
solve using the five step problem solving process, translate, reword, carry-out, check, and state.
An object altitude, in meters is given by the polynomial ht+vt-9.8t^2 where h is the height in meters from which the launch occurs. v is the initial upward speed in meters per second and t is the number of seconds for which the rocket is airborne. A pebble is shot upward from the top of a building 164 meters tall. If the initial speed is 36 meters per second. How high above ground will the peble be after 3 seconds. (round to the nearest tenth of a meter).

ht +vt -9.8t^2

h=164 meters + vt=36(3) -9.8t^2
h=164 meters + 108 -9.8t^2
h=164 meters +172 meters -9.8t^2
answer 163.8t^2 i do not know this but i hope i'm right

I can't claim to understand the problem as you've typed it......

BUT....

I'm pretty sure that "t" is the time in seconds (didn't you say that?.....t is the number of seconds for which the rocket is airborne

There's something seriously missing (or incorrect) from this problem. I know that some of our very dedicated helpers are looking at it, and I hope they'll do better with it than I could ever hope to do.
 


Sorry -- the phone rang.

.
hopen2impov said:
ht + vt - 9.8t^2

h=164 meters + vt=36(3) -9.8t^2 ? Yikes! This notation is a total mess.

Please do not insert English words into the middle of math expressions.

Please do not insert equations like vt=36(3) into an existing expression. That is very confusing for readers. You are not given an equation, so I don't expect to see any equations.

I understand your thought process, but you must not write the steps that you're thinking as part of the math notation!

Also, you properly substituted t = 3 in one term, but you completely missed the number t^2. You need to substitute t = 3 everywhere there is t.

The height of the pebble (above the ground) t seconds after launch is:

h + v*t - 9.8*t^2

This is an expression containing the symbols h, v, and t.

They gave you the value for each of these symbols:

h is 164

v is 36

t is 3

Now, write the expression down, just as it's given, but replace the symbols with their numbers.

164 + 36*3 - 9.8*3^2

Evaluate this expression.

It's that simple.

I don't know why you keep typing ht. Please proofread your posts before submission. Use the [Preview] button.
 
hopen2impov said:
solve using the five step problem solving process, translate, reword, carry-out, check, and state.
An object altitude, in meters is given by the polynomial ht+vt-9.8t^2 where h is the height in meters from which the launch occurs. v is the initial upward speed in meters per second and t is the number of seconds for which the rocket is airborne. A pebble is shot upward from the top of a building 164 meters tall. If the initial speed is 36 meters per second. How high above ground will the peble be after 3 seconds. (round to the nearest tenth of a meter
ht +vt -9.8t^2

h=164 meters + vt=36(3) -9.8t^2
h=164 meters + 108 -9.8t^2
h=164 meters +172 meters -9.8t^2
answer 163.8t^2 i do not know this but i hope i'm right

You know, after spending lots of agonizing minutes with this problem, I'm wondering if the "t" in ht and vt might be a SUBSCRIPT????

As in

h[sub:1z6rg0j8]t[/sub:1z6rg0j8] + v[sub:1z6rg0j8]t[/sub:1z6rg0j8] - 9.8t[sup:1z6rg0j8]2[/sup:1z6rg0j8]


Is h[sub:1z6rg0j8]t[/sub:1z6rg0j8] perhaps the HEIGHT at time t?

And I'm not sure what "v" represents, but maybe v[sub:1z6rg0j8]t[/sub:1z6rg0j8] is the value of v at whatever time "t" represents?

ARGH.
 


I'd like to see subscripted parameters, such that ordered pairs are (t,h).

h = h[sub:3i3ejf75]0[/sub:3i3ejf75] + v[sub:3i3ejf75]0[/sub:3i3ejf75]t - 9.8t^2

I think that's fairly standard now.

(Oh, heck. What am I saying?! I don't really know what the standard is in school anymore.)

 
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