Quadratic applications question

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Homework Help Overview

The problem involves a projectile launched upward, with its height described by the equation -16t² + 320t. Participants are tasked with determining the time it takes for the projectile to return to the ground, specifically solving the equation algebraically.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different algebraic approaches to solving the quadratic equation, with some attempting to simplify the equation by dividing through by -16. Questions arise regarding the validity of the solutions found and the interpretation of the time variable t, particularly at t=0.

Discussion Status

The discussion is active, with participants verifying their calculations and questioning the assumptions about the time variable. Some guidance is offered regarding the interpretation of t=0 as a valid point in the context of the problem, while others express confusion over the calculations and the placement of values in the equation.

Contextual Notes

Participants are grappling with the implications of the solutions found and the physical meaning of the time variable in relation to the projectile's motion. There is an emphasis on ensuring that the solutions align with the physical context of the problem.

ghostchaox
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Homework Statement


A projectile is launched upward. Its height above the ground after t seconds is -16t^2 + 320t. After how many seconds in the air will it return to the ground? Solve algebraically.


Homework Equations


ax^2+bx+c=0 ?
-b (+/-) (Square root of b^2 -4ac)/2a ?


The Attempt at a Solution



t= -320 (+/-) (Square root of 320^2 -4(-16)(0)

t= -320 (+/-) 320/ -16
t= {0,20} ?

If i check it, it doesn't fit.
 
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Its simpler this way.

[itex]0=-16t^2+320t[/itex]

Divide both sides by -16

[itex]0=t^2-20t[/itex]

[itex]0=t(t-20)[/itex]

[tex]t=0,20[/tex]

We get the same solutions, check it the long way by paper, maybe your calculators got errors. My solutions are verified by my calculator..
 
20=-16(20^2) + 320(20) ?
20= -6400+6400?
20=0?

20 =/= 0

0= -16(0^2) + 320(0) ?
0=0+0?
0=0


Now, since this is geometry, t(seconds) cannot equal 0 or else the time and distance would be also 0

If i plug in 20 as t, it ends up as 20=0, which is not true.

I don't understand how I am right
 
t(seconds) cannot equal 0 or else the time and distance would be also 0

Why not? t=0 is just the point of firing - one of two points described by the parabola where the height is zero. What other value of t allows the equation:

-16t^2 + 320t = 0

to equal zero? This:

20=-16(20^2) + 320(20)

Is wrong. Why have you placed 20 on the left hand side?
 
Right... lol Thanks!~!~
 

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