in the calculations shown, what will the units of the final answer be? 18 meters/second x 3600 secons/1 hour x 1 mile/ 1609meter
1 answer:
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The function of the path of the punted ball is a quadratic function which
follows the path of a parabola.
The correct responses are;
Part A: The coordinates of the vertex is
Part B: The maximum height of the punt is <u>36.81 ft.</u>
Part C: The defensive player must reach up to <u>7.65 feet</u> to block the punt.
Part D: The distance down the field the ball will go without being blocked is approximately <u>119.67 ft.</u>
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Reasons:
The function for the height of the punted ball is; h = -0.01·x² + 1.18·x + 2
Assumption; The distances are feet.
Part A: By completing the square, we have;
f(x) = -0.01·x² + 1.18·x + 2
100·f(x) = -x² + 118·x + 200
-100·f(x) = x² - 118·x - 200
x² - 118·x + (118/2)²= 200 + (118/2)²
(x - 59)² = 200 + (59)² = 3681
(x - 59)² - 3681
At the vertex, -3281 = -100·f(x)
∴ f(x) at the vertex = -3681/-100 = 36.81
Part B: The maximum height is given by the y-value at the vertex = 36.81 ft.
Part C: When <em>x</em> = 5, we have;
h = -0.01·x² + 1.18·x + 2
h = -0.01 × 5² + 1.18 × 5 + 2 = 7.65
The defensive player must reach up to 7.65 feet to block the punt
Part D: The distance the ball will go before it hits the ground is given by
the function, for the height as follows;
h = -0.01·x² + 1.18·x + 2 = 0
From the completing the square method, above, we get;
-0.01·x² + 1.18·x + 2 = 0
x² - 118·x - 200 = 0
x² - 118·x + (118/2)²= 200 + (118/2)²
x² - 118·x + (59)²= 200 + (59)² = 3681
(x - 59)² = 3681
x - 59 = ±√3681
x = 59 ± √3681
x = 59 + √3681 ≈ 119.67
The distance down the field the ball will go without being blocked, x ≈ <u>119.67 ft.</u>
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