Let's start by realizing the equation -16t^{2 }+ 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 = -16t^{2} + 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 -16t^{2} + 64t + 80 = 128

This will tell you when the ball will be at 128 feet.

Subtract 128 from both sides and your equation becomes

-16t^{2 }+64t -48 = 0

Factor out a -16 and now your equation is

t^{2} - 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.

Chloe B.

Thank you so much!10/27/20