Question

I'm having a bit of a hard time figuring out this problem and I'm really confused. Could someone check my work so far?
The motion of a ball scooped by a field hockey player can be modeled by h = -16t^2 + 40t, where t is the time in seconds and h is the height of the ball. Will the ball ever reach 22 feet?

22= -16^2t + 40t
22= (4t - 8t)^2 + 40t
(Radical 22)= 4t - 8t + 40t
(Radical 22) = 36t

Answer 1

You started off well, but then it made no more sense

22 = -16t^2 + 40t
16t^2 - 40t + 22 = 0
using the formula, since it does not factor .....
t = (40 ± √(1600 - 4(16)(22)) )/32
= (40 ± √192)/32
= (40 ± 8√3)/32 = (5 ± √3)/4 =appr 1.68 or a negative, which would make no sense.

Answer 2

Wait a minute, I just realized that i missed a step.
22= -16t^2 + 40t
22/-16= -16t^2/-16 + 40t/-16
-11/8= t^2 -10/4t
Now multiply the coefficient of the t-term by 1/2
(- 10/4t)1/2 = -10/8
(-10/8) -10/8 = 100/64 = 25/ 16
-11/8 + 25/16 = t^2 -10/4t +25/16
3/16= (?t -?)^2

Answer 3

Sorry, I hadn't refreshed the page before I sent this.

Answer 4

You are doing "completing the square" in a much-too complicated way:
Consider this after your start:
22= -16t^2 + 40t
16t^2 - 40t = -22
divide by 16
t^2 - (5/2)t = -11/8
t^2 - (5/2)t + 25/16 = -11/8 + 25/16
(t - 5/4)^2 = 3/16
t - 5/4 = ±√3/4
t = 5/4 ± √3/4 <--- same as my answer, and you almost had it

Answer 5

I think I understand it now. Thank you so much for your help.

Answer 6

the maximum of the parabola lies on the axis of symmetry
... t = -b / 2a = -40 / -32 = 1.25

plug that into the height equation to find the max height