- anonymous

Been stuck for hours
The function H(t) = −16t2 + 90t + 50 shows the height H(t), in feet, of a projectile after t seconds. A second object moves in the air along a path represented by g(t) = 28 + 48.8t, where g(t) is the height, in feet, of the object from the ground at time t seconds.
Part A: Create a table using integers 1 through 4 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points)
Part B: Explain what the solution from Part A means in the context of the problem. (4 points)

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- anonymous

- anonymous

Anybody got an idea? I'm basically just needing help on part A.

- mathstudent55

Start by evaluating both functions at t = 1, t = 2, t = 3, t = 4

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## More answers

- anonymous

16 90 28 48?

- anonymous

- mathstudent55

How did you get 16?

- mathstudent55

H(t) = −16t2 + 90t + 50
Let t = 1.
What do you get for H(1) ?

- anonymous

I just found it, idk how i got it, I guessed on it

- anonymous

How do i find h(1)

- anonymous

Like, i'm 100% lost on how I make these, I found some functions but idk how to graph them, can you explain?

- mathstudent55

H(t) = −16t2 + 90t + 50
To find H(1) you replace t with 1, and you evaluate the expression.
H(1) = -16(1)^2 + 90(1) + 50

- mathstudent55

H(1) = -16 + 90 + 50 = 124
Now you need H(2).
Replace t with 2 in the H(t) function and evaluate the right side.
H(2) = -16(2^2) + 90(2) + 50

- anonymous

How do I evaluate it? Like how'd you get that

- anonymous

1 is H(1) = -16 + 90 + 50 = 124, I need to solve for 2,3, and 4?

- anonymous

- anonymous

I need to find 2,3,4 but how?

- mathstudent55

Evaluating an expression is just using arithmetic to calculate what it is equal to.

- mathstudent55

For t = 2, you get
H(2) = -16(2^2) + 90(2) + 50
What is the right side equal to?

- anonymous

H(2) = -16(2v2) + 180 + 50?

- mathstudent55

If I asked you what is 2 + 2, would you answer 2 + 2, or would you answer 4?

- anonymous

H(2) -16^4 + 230?

- mathstudent55

good, keep going

- anonymous

H(2) - 64 + 230?

- mathstudent55

good

- mathstudent55

H(2) = - 64 + 230?

- anonymous

So it'd be H(2) = 294, or since 64 is negative would it be H(2) = 166?

- mathstudent55

H(2) = 166

- mathstudent55

So far we have:
H(1) = 124
H(2) = 166
Now we still need H(3) and H(4) for function H(t)
H(t) = −16t2 + 90t + 50
H(3) = -16(3)^2 + 90(3) + 50

- anonymous

H(3) = -96 + 320

- anonymous

H(3) = 224?

- mathstudent55

No.
First do 3^2

- anonymous

9

- mathstudent55

H(3) = -16(3)^2 + 90(3) + 50
H(3) = -16(9) + 90(3) + 50
Now do -16 * 9. You already have 90(3) + 50 = 320

- anonymous

-144

- anonymous

-144 + 270 + 50 right?

- mathstudent55

yes

- anonymous

H(3)=176?

- mathstudent55

Correct.
Now we need H(4)
H(4) = -16(4)^2 + 90(4) + 50
Remember to start with 4^2 then multiply that by -16

- anonymous

156?

- mathstudent55

I got 154
H(4) = -16(4)^2 + 90(4) + 50 = -256 + 360 + 50 =

- mathstudent55

Ok. We have now function H(t) evaluated at t = 1, t = 2, t = 3, t = 4.
Now we need to do the same for function g(t)

- anonymous

Four more for g(t)?

- anonymous

Alright, what's the functions?

- anonymous

You there math?

- mathstudent55

The g(t) function is simpler. Just multiply the number (1, 2, 3, 4) by 48.8, then add to 28.
g(t) = 28 + 48.8t
g(1) = 28 + 48.8(1) =
g(1) = 28 + 48.8(2) =
g(1) = 28 + 48.8(3) =
g(1) = 28 + 48.8(4) =

- anonymous

g(1) =76.8
g(2)=125.6
g(3)= 174.4
g(4)= 223.2

- mathstudent55

Correct. Great job on function g(t).
Now we place all values in a table like part A asks for.
|dw:1437421534524:dw|

- anonymous

Great, now that the graph's done what now? It says something about what 2 seconds match, or something

- mathstudent55

Part A asks to look at the table and see between which two times the functions H and g have the same value.

- mathstudent55

|dw:1437421963065:dw|

- mathstudent55

Notice that at t = 3, H = 176 and g = 174.4, H is higher,
but at t = 4, H = 154, and g = 223.2, g is higher.
That means at some time between 3 and 4, the functions must have the same value.

- anonymous

I noticed, so that's the answer it requires?

- mathstudent55

The answer to part A is that the objects are at the same height sometime between 3 and 4 seconds.

- mathstudent55

Then the answer to part B is that the objects hit each other at that time between 3 and 4 seconds when their paths cross in the air.

- anonymous

Thank you so much, I honestly would've never figured out how to work this problem. Thanks for the help :))

- mathstudent55

You are welcome.

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