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?

2. Originally Posted by rebak
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?
There must be a typo!

The graph of the given function is a parabola opening down. The vertex of the parabola (that's the highest point the ball can reach!) is at V(0, 10).

Thus: The ball never reaches a height of 40' .

3. You're right, it's actually $\displaystyle 20t-2t^2$ and when it reaches a height of 50

4. 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?

5. 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.

6. No the solution is:

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

7. 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

8. 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

9. Originally Posted by rebak
You're right, it's actually $\displaystyle 20t-2t^2$ and when it reaches a height of 50
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'.

10. Originally Posted by oswaldo
No the solution is:

$\displaystyle 5-\sqrt{5} \; \; and \; \; 5+\sqrt{5}$
No, the question was "For what time interval is the ball higher than 40 feet?" The times the ball is at 40 feet are $\displaystyle 5-\sqrt{5}$ and $\displaystyle 5+\sqrt{5}$. For any time between those two the ball is above 40 feet.

11. Originally Posted by earboth
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'.
Earboth, I understood this to mean that with this new function the ball reaches a maximum height of 50 feet but that the problem was still to find the time interval when the ball was above 40 feet.