Question

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?"

Answer 1

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

Answer 2

@dan815 , @zepdrix

Answer 3

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

Answer 4

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

Answer 5

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 )

Answer 6

My answers are .

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

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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?

Answer 11

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

Answer 12

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?

Answer 13

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).

Answer 14

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.

Answer 15

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?

Answer 16

ah, thanks

Answer 17

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