Question

a ball is thrown directly upward from a height of 3 ft with an initial velocity of 24 ft/sec. the function ( s(t) = -16t^2 + 24t + 3 ) gives the height of the ball, in feet, ( t ) seconds after it has been thrown. determine the time at which the ball reaches its maximum height and find the maximum height. the ball reaches its maximum height of (square) ft (square) sec(s) after the ball is thrown. (type integers or decimals.)

Explanation:

Step1: Identify the function type

The height function \( s(t) = -16t^2 + 24t + 3 \) is a quadratic function in the form \( ax^2 + bx + c \), where \( a = -16 \), \( b = 24 \), \( c = 3 \). For a quadratic function, the vertex (which gives the maximum here since \( a < 0 \)) occurs at \( t = -\frac{b}{2a} \).

Step2: Calculate time at maximum height

Substitute \( a = -16 \) and \( b = 24 \) into the formula:
\[
t = -\frac{24}{2 \times (-16)} = -\frac{24}{-32} = \frac{24}{32} = 0.75
\]

Step3: Calculate maximum height

Substitute \( t = 0.75 \) into \( s(t) \):
\[
s(0.75) = -16(0.75)^2 + 24(0.75) + 3
\]
First, calculate \( (0.75)^2 = 0.5625 \):
\[
-16 \times 0.5625 = -9
\]
\[
24 \times 0.75 = 18
\]
Then, \( s(0.75) = -9 + 18 + 3 = 12 \)

Answer:

The ball reaches its maximum height at \( 0.75 \) sec(s) after the ball is thrown.
The ball reaches its maximum height of \( 12.00 \) ft.