Quadratic applications question

[ghostchaox]

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.

 

[Gib Z]

Its simpler this way.

$0=-16t^2+320t$

Divide both sides by -16

$0=t^2-20t$

$0=t(t-20)$

$$t=0,20$$

We get the same solutions, check it the long way by paper, maybe your calculators got errors. My solutions are verified by my calculator..

 

[ghostchaox]

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

 

[Sojourner01]

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?

 

[ghostchaox]

Right... lol Thanks!~!~