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Compassionate
 one year ago
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
 one year ago
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?"

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Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0Would the equation be \[Height = 16t^2 + 200t + 3\]

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 , @zepdrix

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[h(t) = h_0 + ut  \frac{1}{2}gt^2\]\(u\) is the initial velocity, \(g\) is the acceleration due to gravity (generally taken as \(10 \) m/s^2 = \(32 \) ft/sec^2). \[h(t) = 3 + 200t  16t^2 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1so you were pretty close... since gravity is in the downward direction, we include it with a negative sign.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0Ah, 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
 one year ago
Best ResponseYou've already chosen the best response.0My answers are . A. 628 feet B. 1,878 feet C. 20.87 feet D. 199.33 feet

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1What 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
 one year ago
Best ResponseYou've already chosen the best response.1\[16 \left(\frac{200}{32}\right)^2 + 200\left(\frac{200}{32}\right) + 3\]

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0Wait, so I think I'm starting to get it. When you just solve vertex, you are isolating tunits. When you plug those tunits 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
 one year ago
Best ResponseYou've already chosen the best response.1It's the same thing. I just like to work with fractions. Just about personal preference.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0So, now that I've gotten that cleared up, I have another question. When I isolate my xintercepts in a quadratic equation, am I finding the time it takes xobject to hit the ground, or am I Finding it's velocity?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1A graph consists of lots of points. The xcoordinate of a point is just any value you can input into an expression, and the corresponding output is the ycoordinate of the point. See, a vertex is actually a point. The xcoordinate (b/2a) is the input and the ycoordinate is the output corresponding to it. Now you only got the xcoordinate (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
 one year ago
Best ResponseYou've already chosen the best response.1It depends. In this particular question, you did not find any xintercepts of course. You find the xintercepts 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
 one year ago
Best ResponseYou've already chosen the best response.0dw: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?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Exactly. The negative solution is neglected because it doesn't make sense for time to be negative.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0Mind if you watch me do an example problem or two

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444075119942:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes, sure, go ahead.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0An 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
 one year ago
Best ResponseYou've already chosen the best response.0Sigh, but again, my answers are A. 45 feet B. 13.19 feet C. 36.64 feet D. 95.26 feet

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0\(\frac{45}{32}\) is the time it takes to reach the maximum point

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yes, and you plug that in into the equation

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ b }{ 2(a) } = \frac{ 45 }{ 32 }\]

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0It's negative 45, not positive.

phi
 one year ago
Best ResponseYou've already chosen the best response.0***So, H(t) = 45/2(16) *** it is t= 45/(2*16) = 45/32 t= 45/32 now find H(45/32)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The mistake in your work was that you wrote \(45(1.4)\) as \(45(1.96\) in the next step.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1the answer should definitely be 2025/64 = 31.64...

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0I wrote it as 1.96 because... 16t^2 = 16t(1.4)^2

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1unless of course the golfer is initially standing at a height of 5 feet.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0And 1.4^2 = 1.96

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0h = 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
 one year ago
Best ResponseYou've already chosen the best response.0You should rethink 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)

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.0H(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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