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
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:
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