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A football is punted from a height of 2.5 feet above the ground with an initial velocity of 45 feet per second. A) use the vertical motion model to write an equation that gives the height h in feet as a function of the time t in seconds after it has been punted. B) the football is caught 5.5 feet above the ground. Find the amount of time the football was in the air

User Kemis
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Final answer:

The vertical motion of the football can be modeled using the equation h(t) = -16t^2 + 45t + 2.5. To find the amount of time the football was in the air, we set h(t) equal to 5.5 and solve for t, which gives us a time of about 3.03 seconds.

Step-by-step explanation:

The vertical motion of the football can be modeled using the equation h(t) = -16t^2 + vt + h0, where h(t) is the height of the football at time t, v is the initial velocity, and h0 is the initial height. In this case, the initial height is 2.5 feet and the initial velocity is 45 feet per second, so the equation becomes h(t) = -16t^2 + 45t + 2.5.

To find the amount of time the football was in the air, we need to find the time when the height is 5.5 feet. So we set h(t) equal to 5.5 and solve for t:

5.5 = -16t^2 + 45t + 2.5

Rearranging the equation, we get:

-16t^2 + 45t - 3 = 0

Using the quadratic formula, we can solve for t:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values a = -16, b = 45, and c = -3, we get:

t = (-45 ± √(45^2 - 4(-16)(-3))) / (2(-16))

t ≈ 3.03 seconds or t ≈ 0.12 seconds

Since time cannot be negative, we can conclude that the football was in the air for about 3.03 seconds.

User Thomas Wright
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