Question

a baseball is hit into the air, and its height h in feet after t seconds is given by h(t) = -16t² + 64t + 1. (a) what is the height of the baseball when it is hit? (b) after how many seconds does the baseball reach its maximum height? (c) determine the maximum height of the baseball. (a) the height of the baseball when it is hit is \\(\square\\) ft. (b) the baseball reaches its maximum height after \\(\square\\) sec. (c) the maximum height of the baseball is \\(\square\\) ft.

Explanation:

Step1: Find the height when the baseball is hit (t = 0)

Substitute \(t = 0\) into \(h(t)=-16t^{2}+64t + 1\).
\(h(0)=-16\times0^{2}+64\times0 + 1\)

Step2: Find the time to reach maximum height

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-16\), \(b = 64\)), the time \(t\) at the vertex (maximum for \(a<0\)) is given by \(t=-\frac{b}{2a}\).
\(t=-\frac{64}{2\times(-16)}\)

Step3: Find the maximum height

Substitute the value of \(t\) (from Step2) into \(h(t)\). Let \(t = 2\), then \(h(2)=-16\times2^{2}+64\times2+1\)

Answer:

(a) \(1\) ft.
(b) \(2\) sec.
(c) \(65\) ft.