Answer:
The distance up the ramp in meters she should release the ball if she wants it to reach the bottom in 1.9 seconds is approximately 10 meters
Step-by-step explanation:
The given parameters of the motion of Kathy are expressed in a tabular form after sorting in increasing order using Microsoft Excel, as follows, as follows;
[tex]\begin{array}{cc}t \ (sec)&d \ (meters)\\0.07&1.5\\0.27&1.69\\0.3&2.1\\0.6&2.4\\0.91&3.31\\1.1&4.32\end{array}[/tex]
The general form of the quadratic equation is d = a·t² + b·t + c
Taking the points (0.07, 1.5), (0.27, 1.69), and (0.6, 2.4), which are shown to best follow the path of a quadratic (parabolic) curve by using the Chart function to plot the data on Microsoft Excel, we have;
1.5 = a·0.07² + b·0.07 + c = 0.0049·a + 0.07·b + c
1.69 = a·0.27² + b·0.27 + c = 0.0729·a + 0.27·b + c
2.4 = a·0.6² + b·0.6 + c = 0.36·a + 0.6·b + c
1.5 = 0.0049·a + 0.07·b + c
1.69 = 0.0729·a + 0.27·b + c
2.4 = 0.36·a + 0.6·b + c
Solving with the above system of equations, we have;
a ≈ 2.267
b ≈ 0.179
c ≈ 1.476
Therefore, we have;
d ≈ 2.267·t² + 0.179·t + 1.476
The distance, 'd', up the ramp in meters from which she should release the ball so that the ball reaches the bottom in t = 1.9 seconds is given as follows;
d ≈ 2.267 × (1.9)² + 0.179 × (1.9) + 1.476 = 9.9997
The distance up the ramp in meters she should release the ball if she wants it to reach the bottom in t = 1.9 seconds is d ≈ 9.9997 meters which is approximately 10 meters.
