Question

if a projectile is fired straight upward from the ground with an initial speed of 64 feet per second, then its height h in feet after t seconds is given by the function h(t) = -16t² + 64t. find the maximum height of the projectile. the maximum height of the projectile is (simplify your answer.)

Explanation:

Step1: Identify the vertex of the parabola

The function \( h(t) = -16t^2 + 64t \) is a quadratic function in the form \( y = ax^2 + bx + c \), where \( a = -16 \), \( b = 64 \), and \( c = 0 \). The \( t \)-coordinate of the vertex of a parabola \( y = ax^2 + bx + c \) is given by \( t = -\frac{b}{2a} \).

Substitute \( a = -16 \) and \( b = 64 \) into the formula:
\( t = -\frac{64}{2(-16)} \)
\( t = -\frac{64}{-32} \)
\( t = 2 \)

Step2: Find the maximum height

Now that we have the time \( t = 2 \) seconds when the maximum height occurs, substitute \( t = 2 \) into the height function \( h(t) \):

\( h(2) = -16(2)^2 + 64(2) \)
\( h(2) = -16(4) + 128 \)
\( h(2) = -64 + 128 \)
\( h(2) = 64 \)

Answer:

64