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
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
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.
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
Sorry, I hadn't refreshed the page before I sent this.
I think I understand it now. Thank you so much for your help.
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
To check your work, let's go through the problem step by step.
First, the given equation is h = -16t^2 + 40t, where h represents the height of the ball and t represents time in seconds.
To determine if the ball will reach a height of 22 feet, we need to solve the equation h = 22.
Substituting 22 for h in the equation, we have:
22 = -16t^2 + 40t
Now, we need to rearrange the equation to set it equal to zero. Subtracting 22 from both sides of the equation gives us:
-16t^2 + 40t - 22 = 0
Next, to solve for t, we can use the quadratic formula. The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 40, and c = -22. Plugging these values into the quadratic formula, we have:
t = (-40 ± √(40^2 - 4*(-16)*(-22))) / (2*(-16))
Simplifying further:
t = (-40 ± √(1600 - 1408)) / (-32)
t = (-40 ± √192) / (-32)
Now, we can simplify the expression under the square root:
t = (-40 ± 8√3) / (-32)
Splitting the fraction, we have:
t = -40/(-32) ± 8√3/(-32)
t = 5/4 ± √3/4
So, the two solutions for t are:
t₁ = 5/4 + √3/4
t₂ = 5/4 - √3/4
These two values represent the times at which the ball reaches a height of 22 feet. Consequently, the ball will reach a height of 22 feet at those specific times.
I hope this explanation helps you understand the problem better!