A football is punted from a height of 8 feet above the ground with an initial vertical velocity of 44 feet per second. Write an
equation to model the height h, in feet, of the ball t seconds after it has been punted. The football is caught at 15 feet above the ground. How long was the football in the air?
1 answer:
Assuming ballistic motion, the equation for height as a function of time is
h(t) = -16t² + v0·t + h0
where v0 is the initial vertical velocity (44 ft/s) and h0 is the initial height (8 ft).
You want to find t such that h(t) = 15. Substituting the given values, we have
15 = -16t² +44t +8
16t² -44t +7 = 0
Using the quadratic formula, we find t to be ...
t = (44 ±√(44² - 4·16·7))/(2·16)
t = (11 ± √93)/8
We are not interested in the time when the football is rising through the 15 ft height, so the time of interest is
t = (11 +√93)/8 ≈ 2.580 . . . . seconds
h(t) = -16t² + v0·t + h0
where v0 is the initial vertical velocity (44 ft/s) and h0 is the initial height (8 ft).
You want to find t such that h(t) = 15. Substituting the given values, we have
15 = -16t² +44t +8
16t² -44t +7 = 0
Using the quadratic formula, we find t to be ...
t = (44 ±√(44² - 4·16·7))/(2·16)
t = (11 ± √93)/8
We are not interested in the time when the football is rising through the 15 ft height, so the time of interest is
t = (11 +√93)/8 ≈ 2.580 . . . . seconds
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Answer:
y = x - 2
Step-by-step explanation:
<u>The answer to this problem is a simple plug-in of the given values into the slope-intercept formula.</u>
Slope intercept formula: y = mx + b
(m = slope)
(b = y intercept)
If the equation has a slope of m = , then the slope-intercept form would be:
y = x
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y = mx - 2
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