Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
The parabolic motion is an illustration of a quadratic function
The equation that models that path of the rocket is y = -16/31x^2 + 256/31x - 880/31
How to model the function?
Given that:
x stands for time and y stands for height in feet
So, we have the following coordinate points
(x,y) = (5,0), (11,0) and (10,80)
A parabolic motion is represented as:
y =ax^2 + bx + c
At (5,0), we have:
25a + 5b + c = 0
c= -25a - 5b
At (11,0), we have:
121a + 11b + c = 0
Substitute c= -25a - 5b
121a + 11b -25a - 5b = 0
Simpify
96a + 6b = 0
At (10,80), we have:
100a + 10b + c = 80
Substitute c= -25a - 5b
100a + 10b - 25a -5b = 80
75a -5b = 80
Divide through by 5
15a -b = 16
Make b the subject
b = 15a + 16
Substitute b = 15a + 16 in 96a + 6b = 0
96a + 6(15a + 16) = 0
Expand
96a + 90a + 96 = 0
This gives
186a = -96
Solve for a
a = -16/31
Recall that:
b = 15a + 16
So, we have:
b = -15 * 16/31 + 16
b =-240/31 + 16
Take LCM
b =(-240 + 31 * 16)/31
[tex]b =256/31
Also, we have:
c= -25a - 5b
This gives
c= 25*16/31 - 5 * 256/31
Take LCM
c= -880/31
Recall that:
y =ax^2 + bx + c
This gives
y = -16/31x^2 + 256/31x - 880/31
Hence, the equation that models that path of the rocket is y = -16/31x^2 + 256/31x - 880/31
Read more about parabolic motion at:
https://brainly.com/question/1130127
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.