Let's start by realizing the equation -16t2 + 64t + 80t is incorrect.
The equation should read -16t + 64t + 80.
There are 2 ways you can work this problem. Decide which is easier for you and just go with it. I will walk you through both.
First, looking at your answer choice, it is obvious that A and B are incorrect. Time can never be negative. Therefore you are only left with C or D.
The easiest way to answer this question is to find the vertex of the parabola.
You remember from Algebra 1.
The vertex (x,y) is found with the following:
X = -b/2a. a=-16, b=64, c=80
X = -64 / 2(-16) = -64 / -32 = 2
Substitute 2 for t in your equation = -16t2 + 64t +80 =
-16(4) + 64(2) + 80 =
-64 + 128 + 80 = 144
144 is the y in your vertex. So the vertex is (2, 144).
This means that in 2 seconds the height of the ball will be at 144 feet (the highest height it will reach.
So, since 144 > 128, 2 seconds must be within the interval of time.
Between C and D, only D allows for 2 to be in your answer.
Therefore, 1< t < 3.
You can also workout the equation -16t2 + 64t + 80 = 128
This will tell you when the ball will be at 128 feet.
Subtract 128 from both sides and your equation becomes
-16t2 +64t -48 = 0
Factor out a -16 and now your equation is
t2 - 4t + 3 = 0
(Remember to change your signs when you divide by a negative)
Factoring the equation we get (t - 1)(t - 3) = 0
Therefore t = 1 or t = 3.
This tells us that at 1 second and at 3 seconds the ball will be AT 128 feet.
At 1 second, the ball rises to 128 feet. At 3 seconds, the ball is on its way back down, and will reach 128 feet once again.
Therefore, the ball is higher than 128 feet between 1 and 3 seconds.
1 < t < 3.
Hope this helps.