For this function $\displaystyle h(t)=10-4t^2$ how do you determine what time interval the ball will be more than 40 feet in the air?

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- Feb 16th 2009, 08:59 PMrebakQuadratic Model
For this function $\displaystyle h(t)=10-4t^2$ how do you determine what time interval the ball will be more than 40 feet in the air?

- Feb 16th 2009, 11:37 PMearboth
- Feb 17th 2009, 05:57 AMrebak
You're right, it's actually $\displaystyle 20t-2t^2$ and when it reaches a height of 50

- Feb 17th 2009, 07:26 AMoswaldo
I am assuming t is time. As said max height is 50 at t=5. Due to symmetry of the curve it should be [5-a, 5+a].

20t-2t^2=40 gives you the solution.

I think it is: [5-sqrt(5), 5+sqrt(5)].

-O

P.S. I knew that plot of an object is a parabolla, f(x)=ax^2+bx+c, where x is the horizontal range. BUT I am not sure if it is still a parobolla when it is written with respect to time: f(t). Would that trajectory happen on the Jupiter? :) - Feb 17th 2009, 07:42 AMrebak
So is that -5 and 5? That's what I got originally, but I thought it didn't make sense to have the interval start at a negative number.

- Feb 17th 2009, 07:46 AMoswaldo
No the solution is:

$\displaystyle

5-\sqrt{5} \; \; and \; \; 5+\sqrt{5}

$ - Feb 17th 2009, 09:43 AMrebak
Thanks for your help! Did you use the quadratic formula to solve it? I used factoring and it seemed to work fine, so I wasn't sure where the square roots came from

- Feb 17th 2009, 09:48 AMoswaldo
20t-2t^2=40

-2t^2+20t-40=0

t^2 -10t - 20=0

disc.= 10^2 - 4*1*(-20)=20

t1= (10-sqrt(20))/2

t2= (10+sqrt(20))/2

gives you the answer. You can factorize as well, but the above is easier.

-O - Feb 17th 2009, 10:29 AMearboth
If I understand you correctly you have to solve

$\displaystyle 20t-2t^2 \geq 50~\implies~2t^2-20t+50\leq 0~\implies~t^2-10t+25 \leq 0~\implies~(t-5)^2\leq 0$

The LHS of the inequality is a square which never will be negative (or smaller than zero - that's the same). At t = 5 the LHS will be zero.

Thus the answer is: At t = 5 the ball reaches a height of 50'. - Feb 17th 2009, 11:41 AMHallsofIvy
- Feb 17th 2009, 11:46 AMHallsofIvy