The equation of an object in free fall is the equation of a parabola
- The equation for the path of the ball is y = -16·x² + 36·x + 4
- The baseball height, after 1.7 seconds is 18.89 feet.
The reason the above values are correct are as follows:
The known parameters are;
The direction of the baseball is up in the air
The values of the height of the baseball after x seconds is given as follows;
[tex]\begin{array}{ccc}\mathbf{Time, \ x}&&\mathbf{Baseball \ height, \ y}\\0.5&&18\\1&&24\\1.5&&22\\2&&12\end{array}\right][/tex]
To write the equation for the path of the baseball
Solution:
The general equation of the motion of free fall is the equation of a parabola, which can be written as follows;
y = a·x² + b·x + c
From the given table, we have;
18 = a·0.5² + b·0.5 + c = 0.25·a + 0.5·b + c
18 = 0.25·a + 0.5·b + c...(1)
24 = a + b + c...(2)
22 = 2.25·a + 1.5·b + c...(3)
12 = 4·a + 2·b + c...(4)
Subtracting equation (1), from (2), to get equation (5) and subtracting equation (3) from (4) to get equation (6) gives;
6 = 0.75·a + 0.5·b...(5)
-10 = 1.75·a + 0.5·b...(6)
Subtracting equation (5) from equation (6) gives;
-10 - 6 = 1.75·a - 0.75·a = a
a = -16
From equation (5), we get;
6 = 0.75 × (-16) + 0.5·b
b = (6 - 0.75 × (-16))/0.5 = 36
b = 36
From equation (2), we get;
24 = (-16) + 36 + c
c = 24 - ((-16) + 36) = 4
The equation for the path of the ball is y = -16·x² + 36·x + 4
The height of the baseball after 1.7 seconds
Solution:
The height, y, of baseball after 1.7 seconds is given by substituting for the time x = 1.7 in the equation of the baseball as follows;
After 1.7 seconds, y = -16 × 1.7² + 36 × 1.7 + 4 = 18.89
The height of the baseball y, after 1.7 seconds is 18.89 feet.
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