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An object is launched directly in the air at a speed of 16 feet per second from platform located 7 feet above the ground. The position of the object can be modeled using the function
f(t)=-16^2+16t+7, where t is the time in seconds and f(t) is the height, in feet, of the object. What is the maximum height, in feet, that the object will reach?

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Explanation:

To find the maximum height of the object, we need to determine the vertex of the parabolic function f(t) = 16t^2 + 16t + 7.

The vertex of a parabola in the form f(t) = at^2 + bt + c is given by (-b/2a, f(-b/2a)). In this case, a = 16, b = 16, and c = 7, so the vertex is:

(-b/2a, f(-b/2a)) = (-16/(216), f(-16/(216))) = (-1/2, f(-1/2))

To find f(-1/2), we can substitute t = -1/2 into the function:

f(-1/2) = 16(-1/2)^2 + 16(-1/2) + 7 = 8 - 8 + 7 = 7

Therefore, the vertex is at (-1/2, 7). This means that the maximum height of the object is 7 feet above the ground.

Alternatively, we could also use the fact that the maximum height occurs at the vertex of the parabola, which is the point where the derivative of the function is zero. The derivative of f(t) is:

f'(t) = 32t + 16

Setting this equal to zero and solving for t, we get:

32t + 16 = 0

t = -1/2

So the maximum height occurs at t = -1/2, which corresponds to the vertex of the parabola, and the maximum height is 7 feet.

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