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Mathematics 85 Online
OpenStudy (anonymous):

The function H(t) = -16t^2 + vt + s shows the height H (t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second. Part A: The projectile was launched from a height of 100 feet with an initial velocity of 60 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. Part B: What is the maximum height that the projectile will reach? Show your work

OpenStudy (anonymous):

Part C: Another object moves in the air along the path of g(t) = 20 + 38.7t where g(t) is the height, in feet, of the object from the ground at time t seconds. Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values]

OpenStudy (anonymous):

Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?

OpenStudy (anonymous):

@mathmale @triciaal @mathstudent55 @sleepyjess @ganeshie8 @zepdrix

OpenStudy (misty1212):

HI!!

OpenStudy (anonymous):

Hi

OpenStudy (misty1212):

\[H(t) = -16t^2 + vt + s\] put \(v=60,h=100\) to get your function

OpenStudy (misty1212):

that takes care of part A

OpenStudy (misty1212):

do you know how to find the vertex?

OpenStudy (anonymous):

\[h(t) = -16t^2 + 60t + 100?\]

OpenStudy (misty1212):

yes that is it

OpenStudy (anonymous):

Yes, -b/2a to find x coordinate. Input x coordinate in function to find y

OpenStudy (misty1212):

i meant put \(x=100\) but you got it right

OpenStudy (misty1212):

yes that is right

OpenStudy (misty1212):

looks like you are in good shape

OpenStudy (anonymous):

@misty1212 shouldn't s = 100?

OpenStudy (boldjon):

You already have the equation: 0 = -16*t² + 80*t + 96; a quadratic in t. Use the quadratic formula to find t.

OpenStudy (misty1212):

yes i said it wrong twice

OpenStudy (anonymous):

Hold on, my internet is freaking out

OpenStudy (misty1212):

\(s=100\) your function is correct

OpenStudy (anonymous):

@boldjon Do I need to find t for any of the parts? i have no idea lol

OpenStudy (misty1212):

the function @boldjon has must be from some other question, not this one

OpenStudy (anonymous):

\[x = \frac{ -b \pm \sqrt(b^2 - 4ac) }{ 2 }\]

OpenStudy (anonymous):

wait, that's wrong

OpenStudy (misty1212):

yes it is, but you do not need that, that is for where \(h=0\) i.e when it hits the ground, which i don't think you are asked

OpenStudy (anonymous):

\[x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }\]

OpenStudy (anonymous):

alright

OpenStudy (anonymous):

So Part A is done as we have already created the equation. Part B i can probably solve on my own as I know it's -b/2a and then input the x coordinate into the function

OpenStudy (anonymous):

I'll do part b anyways though Part B: What is the maximum height that the projectile will reach? Show your work

OpenStudy (anonymous):

H(t) = -16t^2 + 60t + 100 I believe this is written in ax^2 + bx + c form, correct @misty1212 ?

OpenStudy (misty1212):

eys

OpenStudy (anonymous):

alright, so a = 16, b = 60, and c = 100

OpenStudy (misty1212):

no \(a=-16\)

OpenStudy (anonymous):

oh right

OpenStudy (anonymous):

-b/2a = -60/2*-16 2 * -16 = -32 -60 / -32 = 1.87500 Which can be rounded down to 1.88

OpenStudy (misty1212):

i would leave it as a fraction, reduce it, not convert to a decimal

OpenStudy (anonymous):

alright

OpenStudy (misty1212):

\[\frac{15}{8}\]

OpenStudy (misty1212):

or go ahead and use \(1.875\) if you like, you will probably be best off with a calculator anyways

OpenStudy (anonymous):

I work better with decimals, can we do the exact decimal form instead?

OpenStudy (anonymous):

alright, thanks

OpenStudy (misty1212):

yeah go ahead and use \(1.875\)

OpenStudy (misty1212):

i can check what you get if you would like

OpenStudy (anonymous):

H(1.875) = -16(1.875)^2 + 60(1.875) + 100 Solve as usual 1.875^2 = 3.515625 3.515625 * -16 = -56.25. 60 * 1.875 = 112.5 Add together -56.25 + 112.5 + 100 = 156.25

OpenStudy (misty1212):

yup easy if you use this http://www.wolframalpha.com/input/?i=-16%281.875%29%5E2%2B60*1.875%2B100

OpenStudy (anonymous):

That's strange, I input the equation into desmos and it says it's (-1.88, 43.75) @misty1212 any ideas?

OpenStudy (misty1212):

no but you and wolfram are right probably something with parentheses

OpenStudy (misty1212):

oh you should use \(1.875\) not \(-1.88\) 'the sign is wrong there

OpenStudy (anonymous):

Now it's correct. For some reason desmos didn't understand h(t), probably a bug. I switched all the t's with x's and now it says the correct vertex

OpenStudy (anonymous):

(1.88, 156.25)

OpenStudy (anonymous):

@misty1212 That solves Part B Now Part C Part C: Another object moves in the air along the path of g(t) = 20 + 38.7t where g(t) is the height, in feet, of the object from the ground at time t seconds. Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values]

OpenStudy (anonymous):

I think we can use the draw tool for this

OpenStudy (misty1212):

i guess it wants you to check with whole numbers

OpenStudy (misty1212):

i would just go ahead and set them equal and solve if you can use the computer, do this http://www.wolframalpha.com/input/?i=-16x%5E2%2B60x%2B100%3D20%2B38.7x

OpenStudy (anonymous):

Right, I think I can do Part C in a word docuement anyways

OpenStudy (anonymous):

Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?

OpenStudy (misty1212):

did you find when they intersect? you need that first

OpenStudy (anonymous):

2 places when x is abotu -1.66 and when x is about 2.99

OpenStudy (misty1212):

they want you to use whole numbers here the first one does not make any sense, it was in the past 2.99 rounds to 3 they meet in 3 seconds

OpenStudy (misty1212):

as for going up or going down, is 3 seconds before or after the projectile reaches its maximum height?

OpenStudy (anonymous):

I believe after, correct?

OpenStudy (misty1212):

yes so it is going down

OpenStudy (anonymous):

Alright, thanks!

OpenStudy (anonymous):

@misty1212 I am having trouble. Does g(t) even have a maximum height?

OpenStudy (anonymous):

@misty1212 Never mind, I'm good. Thanks for all your help on this!

OpenStudy (misty1212):

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