Quadratic formula: @Zale101
"A pistol is accidentally discharged vertically upward at a height of 3 feet above the ground. If the bullet has an initial muzzle velocity of 200 feet per second, what maximum height will it reach before it starts to fall to the ground?"

- Compassionate

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- Compassionate

Would the equation be \[Height = 16t^2 + 200t + 3\]

- Compassionate

@dan815 , @zepdrix

- ParthKohli

\[h(t) = h_0 + ut - \frac{1}{2}gt^2\]\(u\) is the initial velocity, \(g\) is the acceleration due to gravity (generally taken as \(-10 \) m/s^2 = \(-32 \) ft/sec^2). \[h(t) = 3 + 200t - 16t^2 \]

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## More answers

- ParthKohli

so you were pretty close... since gravity is in the downward direction, we include it with a negative sign.

- Compassionate

Ah, I see, acceleration due to gravity is negative because of it's motion. In order to get the vertex of the parabola, I could simply use the vertex formula \[\frac{ -b }{ 2a }\] So, given the equation \(height = -16t^2 - 200t + 3\) I could say; \[\frac{ -(-200) }{ 2(-16) } = \frac{ 200 }{ -32 }\] I get -6.25, or 6.25, which can't be right because that isn't an answer (I'm using a practice problem from http://www.algebralab.org/Word/Word.aspx?file=Algebra_MaxMinProjectiles.xml )

- Compassionate

My answers are .
A. 628 feet
B. 1,878 feet
C. 20.87 feet
D. 199.33 feet

- Compassionate

I see I made a mistake and did -200, but even if that was corrected, it still wouldn't be anything close to the answers

- ParthKohli

What you got was correct. However, you got the value of \(t\) which makes \(h\) maximum. And you are to calculate the corresponding value of \(h\) for the \(t\) you got (because it asks for the maximum height itself, and not the time at which it is at the maximum height). How would you do that? Of course, you replace \(t\) with \(200/32\) in the quadratic expression which would give you your answer.

- ParthKohli

\[-16 \left(\frac{200}{32}\right)^2 + 200\left(\frac{200}{32}\right) + 3\]

- Compassionate

Wait, so I think I'm starting to get it.
When you just solve vertex, you are isolating t-units. When you plug those t-units in, you can calculate height by just using basic PEMDAS. But, I got a question
Why plug in 200/32. Why not 6.25, wouldn't working with decimals be a tad bit easier?

- ParthKohli

It's the same thing. I just like to work with fractions. Just about personal preference.

- Compassionate

So, now that I've gotten that cleared up, I have another question. When I isolate my x-intercepts in a quadratic equation, am I finding the time it takes x-object to hit the ground, or am I Finding it's velocity?

- ParthKohli

A graph consists of lots of points. The x-coordinate of a point is just any value you can input into an expression, and the corresponding output is the y-coordinate of the point.
See, a vertex is actually a point. The x-coordinate (-b/2a) is the input and the y-coordinate is the output corresponding to it. Now you only got the x-coordinate (input, i.e., time) but you need to output corresponding to that. So you plug that particular time you got (-b/2a) into the expression to get the corresponding output (height).

- ParthKohli

It depends. In this particular question, you did not find any x-intercepts of course. You find the x-intercepts by equating the quadratic expression to zero. If you want to find the time when the object returns to the ground, you take its height function \(h(t) = -16t^2 + 200t + 3\) and equate it to zero.\[-16t^2 + 200t + 3 = 0\]If you solve this equation using the quadratic formula or factorisation or any other method, you'd get the time at which its height becomes zero, i.e., it returns to the ground.

- Compassionate

|dw:1444074992922:dw| I'm talking more about these. IF I take that equation above and plug it into the quadratic and solve for both positive and negative points, the poisitive one will be the time it takes the ball to land, right?

- Compassionate

ah, thanks

- ParthKohli

Exactly. The negative solution is neglected because it doesn't make sense for time to be negative.

- Compassionate

Mind if you watch me do an example problem or two

- ParthKohli

|dw:1444075119942:dw|

- ParthKohli

Yes, sure, go ahead.

- Compassionate

An over zealous golfer hits a flop shot with a sand wedge to get out of the corner of a sand trip with an initial velocity of 45 feet per second. What is the maximum height that the golf ball will reach?
So, \(h = -16t^2 + 45 + 0\) I say the initial height = 0 because he's on the ground
To find the maximum height, I would first need to find the time of the maximimum height
So, H(t) = -45/2(16)
H(t) = 1.4 or -45/32
\(h = -16(1.4)^2 + 45(1.4)\)
\(h = -16(1.96) + 45(1.96)
\(h = -31.36 + 88.2\)
\(h = 56.84\)

- Compassionate

Sigh, but again, my answers are
A. 45 feet
B. 13.19 feet
C. 36.64 feet
D. 95.26 feet

- ParthKohli

http://www.wolframalpha.com/input/?i=45t+-+16t%5E2+maximum

- Compassionate

\(\frac{45}{32}\) is the time it takes to reach the maximum point

- ParthKohli

yes, and you plug that in into the equation

- Compassionate

\[\frac{ -b }{ 2(a) } = \frac{ -45 }{ 32 }\]

- Compassionate

It's negative 45, not positive.

- phi

***So, H(t) = -45/2(16) ***
it is t= -45/(-2*16) = 45/32
t= 45/32
now find H(45/32)

- ParthKohli

The mistake in your work was that you wrote \(45(1.4)\) as \(45(1.96\) in the next step.

- phi

or - (-45)/32

- ParthKohli

the answer should definitely be 2025/64 = 31.64...

- Compassionate

I wrote it as 1.96 because...
-16t^2 = -16t(1.4)^2

- ParthKohli

unless of course the golfer is initially standing at a height of 5 feet.

- Compassionate

And 1.4^2 = 1.96

- Compassionate

h = -16t^2 + 45 + 0
So what do I do with the t^2... Would it be... \[-16(\frac{ 45 }{ 32 })^2 = -16\frac{ 2025 }{ 1024 }\] Which one is it?

- phi

You should re-think that.
the equation is
H(t) = -16 t^2 + 45 t (notice it is 45 t not just 45)
There is only 1 vertex (i.e. peak)
you find the time of the peak using
t= -b/2a
(you only get one number, it will be t= 45/32)
Now find the height
-16*(45/32)^2 + 45*(45/32)

- ParthKohli

h = -16t^2 + 45t

- Compassionate

H(t) = -16 t^2 + 45 t
So, when I scared it, I only square 45? This is confusing. I have to pickup my girlfriend. I'll be back on later.

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