To find the function that models the ball's distance from the ground with respect to time, we can use the points provided to find the coefficients of a quadratic function.
Let's assume the function is of the form:
d(t) = at^2 + bt + c
By plugging in the given points, we get the following system of equations:
1) d(0) = c = 0
2) d(0.25) = 0.25^2a + 0.25b = 2.76
3) d(0.50) = 0.50^2a + 0.50b = 4.90
4) d(0.75) = 0.75^2a + 0.75b = 6.43
5) d(1.00) = a + b = 7.35
6) d(1.25) = 1.25^2a + 1.25b = 7.66
From equation 1, we find that c = 0. Next, we solve this system of equations to get the values of a, b, and c:
From equation 2:
0.25^2a + 0.25b = 2.76
0.0625a + 0.25b = 2.76
From equation 3:
0.50^2a + 0.50b = 4.90
0.25a + 0.50b = 4.90
Subtracting these equations gives:
0.0625a - 0.25a = 2.76 - 4.90
-0.1875a = -2.14
a = 11.4133
Now, we can solve for b:
b = 7.35 - a
b = 7.35 - 11.4133
b = -4.0633
Therefore, the function that models the ball's distance from the ground with respect to time is:
d(t) = 11.4133t^2 - 4.0633t
A ball is placed in a machine that throws the ball up in the air. The table represents some points on the graph of a function that models the ball's distance from the ground with respect to the time since the ball has been thrown.
Time Since Thrown from Machine (seconds)
0
0.25
0.50
0.75
1.00
1.25
Distance from Ground (meters)
0
2.76
4.90
6.43
7.35
7.66
1 answer
1 answer