Precalculus: Vertex Form and etc.

The height of a football is given by the formula y = -16t ^ 2 + 64t + 3
(t is in seconds)

Now i've placed this quadratic in vertex form and arrived at the answer
-16(t - 2) ^ 2 + 67


Now the question i have is at what time is the football at 50 feet (to the nearest tenth of a second)

I did ...


-16(t - 2) ^ 2 + 67 = 50

-67 from both sides


so .. -16(t - 2) ^ 2 = -17

divide 16 by both sides

(t - 2) ^ 2 = -17/-16

or just

(t-2) ^ 2 = 17/16


take the square root of each side, therefore ...

t - 2 = + - the square root of 17/16

if i add 2 to both sides

t = 2 +- the sq. rt. of 17/16

t = 2 + sq. rt. of 17/16 = 3.03 seconds

t = 2 - sq. rt. of 17/16 = .9692 seconds


Now when i'm on my TI-83 and i do second graph i get that when x is 3.03, y = 50.026 .. therefore at 3.03 seconds the football is at 50 feet

is there any other way in words to justify this reason

any help would be great =)

Looks good, well done. I'm not sure what your asking.

Ball is a 50 ft when:


t = 2 + (sqrt (17)/4) = (8 + sqrt(17))/4

and

t = 2 - (sqrt (17)/4) = (8 - sqrt(17))/4


Both answers are correct.

Just write out all your steps, no need for any words to justify it. Pobably a good idea to give both your answers to the same, and appropriate, degree of accuracy. The problem statement ask for t to the nearest tenth of a second.

thank you so much !

therefore when the football is at 50 feet

the times are both:

t = 2 + sq. rt. of 17/ sq. rt. of 16 ( or 2 + sq rt. of 17 / 4 ) = 3.0307

so to the nearest tenth = 3.0

and

t = 2 - sq. rt. of 17/ sq. rt. of 16 (or 2 - sq. rt. of 17 / 4) = .9692

so to the nearest tenth = .97 -> 1.0

Another quick question ..

if a question is asking how long the ball is up in the air for

is that the equivalent of putting the equation i got equal to 0 ..which is

-16(t - 2) ^ 2 + 67 = 0

?

Yes to both.

Don't be confused by the negative root, the ball was not on the ground at t = 0

The positive root gives you the time taken for the ball to hit the ground.

therefore ..

-16(t - 2) ^ 2 + 67 = 0

(t - 2) ^ 2 = 67/16

Take the Sq. Rt. of both sides which equals :

t - 2 = Sq. Rt. of 67/ Sq. Rt. of 16

t = 2 + - Sq. Rt. of 67 / Sq. Rt. of 16

t = 2 + Sq. Rt. of 67/ Sq. Rt. of 16 = 4.046

t = 2 - Sq. Rt. of 67/ Sq. Rt. of 16 = -.046

Therefore is saying the ball is in the air for 4.046 seconds (t = 4.046) correct



?

is that correct?