
Jordan K. answered 10/24/15
Tutor
4.9
(79)
Nationally Certified Math Teacher (grades 6 through 12)
Hi Haley,
Before we get to solving for the time asked in our problem, let's first determine the key information concerning the flight path of the toy rocket.
Let's begin by writing the Height Function, h(t), in standard quadratic form (ax2 + bx + c):
h(t) = -16t2 + 64t (a = -16; b = 64; c = 0)
Next, let's use the axis of symmetry formula (x = -b/2a) for a parabola to determine the time it takes for the toy rocket to reach its maximum height:
x = -b/2a
x = -64/[2(-16)]
x = 64/32
x = 2 seconds to reach maximum height
Next, let's plug in the value of 2 for t in the Height Function, h(t), to determine the maximum height of the toy rocket:
h(t) = -16t2 + 64t
h(t) = -16(2)2 + 64(2)
h(t) = -16(4) + 128
h(t) = -64 + 128
h(t) = 64 feet (maximum height of toy rocket)
Next, let's determine the total flight time of the toy rocket by setting the Height Function, h(t), to zero and solving for time (t):
h(t) = -16t2 + 64t
-16t2 + 64t = 0
16t2 = 64t
t = 64t/16t
t = 4 seconds (total flight time of toy rocket)
Now let's determine between the time interval that the height of the toy rocket is above 48 feet by setting the Height Function, h(t), to 48 and solving for time (t):
h(t) = -16t2 + 64t
-16t2 + 64t = 48
(-16t2)/-16 + 64t/-16 = 48/-16
t2 - 4t = -3
t2 - 4t + 3 = 0
(t - 1)(t - 3) = 0
t = 1 second (height = 48 feet); t = 3 seconds (height = 48 feet)
Therefore, the total time (t) that the height of the toy rocket is above 48 feet:
t < |1-3|
t < |-2|
t < 2 seconds height of toy rocket is above 48 feet
Below is the link to our graph of the flight path of the toy rocket:
https://dl.dropbox.com/s/doyd0849bg1i7wd/Graph_of_Toy_Rocket_Flight.png?raw=1
The points marked in red confirm the same values determined above in our algebraic solution.
Thanks for submitting this problem and glad to help.
God bless, Jordan (Romans 5:8)