Answer:
y = -15.9x² + 110.48x + 3.02
initial velocity: 110.48 ft/s
initial height: 3.02 ft
time to cath the ball at 6 ft: 6.92
Step-by-step explanation:
In the picture attached, the plot of the data points and the equation of the best regression is shown, which correspond to a quadratic equation. (It was obtained using Excel)
The initial velocity is the second coefficient.
The initial height is the third coefficient.
Replacing y = 6 into the equation and solving with the quadratic formula:
6 = -15.9x² + 110.48x + 3.02
0 = -15.9x² + 110.48x + 3.02 - 6
0 = -15.9x² + 110.48x - 2.98
[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/tex]
[tex]x = \frac{-110.48 \pm \sqrt{110.48^2 - 4(-15.9)(-2.98)}}{2(-15.9)} [/tex]
[tex]x = \frac{-110.48 \pm 109.62}{-31.8} [/tex]
[tex]x_1 = \frac{-110.48 + 109.62}{-31.8} [/tex]
[tex]x_1 = 0.03[/tex]
[tex]x_2 = \frac{-110.48 - 109.62}{-31.8} [/tex]
[tex]x_2 = 6.92[/tex]
In the context of the problem, only x2 has sense.