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 function type

The height function \( h(t) = -16t^2 + 64t \) is a quadratic function in the form \( y = ax^2 + bx + c \), where \( a = -16 \), \( b = 64 \), and \( c = 0 \). Since \( a < 0 \), the parabola opens downward, and the vertex represents the maximum point.

Step2: Find the time at maximum height

The \( t \)-coordinate of the vertex of a quadratic function \( y = ax^2 + bx + c \) is given by \( t = -\frac{b}{2a} \). Substituting \( a = -16 \) and \( b = 64 \):
\[
t = -\frac{64}{2(-16)} = -\frac{64}{-32} = 2
\]

Step3: Find the maximum height

Substitute \( t = 2 \) into the height function \( h(t) \):
\[
h(2) = -16(2)^2 + 64(2) = -16(4) + 128 = -64 + 128 = 64
\]

Answer:

64