Get comprehensive solutions to your problems with IDNLearn.com. Find accurate and detailed answers to your questions from our experienced and dedicated community members.
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
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.