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Ryan throws a tennis ball straight up into the air. The ball reaches its maximum height at 2 seconds. The approximate height of the ball $x$ seconds after being thrown is shown in the table.
Motion of Tennis Ball
\begin{tabular}{|c|c|}
\hline \begin{tabular}{c}
Time (seconds) \\
$x$
\end{tabular} & \begin{tabular}{c}
Height (feet) \\
$y$
\end{tabular} \\
\hline 0 & 4 \\
\hline 1 & 52 \\
\hline 2 & 68 \\
\hline 3 & 52 \\
\hline 4 & 4 \\
\hline
\end{tabular}

Which equation models the motion of the ball?
$y=-17(x)(x-4)$
$y=-16(x)(x-4)$
$y=-16(x-2)^2+68$
$y=-17(x-2)^2+68$

Sagot :

Sqdancefanidn

Answer:

  (c)  y = -16(x -2)² +68

Step-by-step explanation:

You want the equation that models the table ...

  (x, y) = (0, 4), (1, 52), (2, 68), (3, 52), (4, 4).

Vertex

The table indicates the maximum height is 68 ft at x = 2 seconds. The equation in vertex form will look like ...

  y = a(x -h)² +k . . . . . . . . . quadratic with vertex (h, k)

  y = a(x -2)² +68 . . . . . . . . quadratic with vertex (2, 68)

Scale factor

We can find the value of 'a' by considering a point where x differs from 2 by 1 unit. (3, 52) is such a point.

  52 = a(3 -2)² +68

  -16 = a

Then the complete equation of motion is ...

  y = -16(x -2)² +68 . . . . . . matches choice C

__

Additional comment

This is consistent with the usual ballistic motion equation ...

  y = -16x² +v₀x

which has -16 as its leading coefficient.

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