Question

a soccer ball is kicked from the ground with an initial upward velocity of 160 feet per second. the equation ( h(t) = -16t^2 + 160t ) gives the height ( h ) of the ball after ( t ) seconds.

a) find the maximum height of the ball.

( h_{\text{max}} = ) (square) feet

b) how may seconds will it take for the ball to reach the ground?

( t = ) (square) seconds

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Explanation:

Part (a)

Step1: Identify the vertex of the parabola

The height function \( h(t) = -16t^2 + 160t \) is a quadratic function in the form \( y = ax^2 + bx + c \), where \( a = -16 \), \( b = 160 \), and \( c = 0 \). The \( t \)-coordinate of the vertex (which gives the time at maximum height) is given by \( t = -\frac{b}{2a} \).
\[
t = -\frac{160}{2(-16)} = -\frac{160}{-32} = 5
\]

Step2: Find the maximum height

Substitute \( t = 5 \) into the height function \( h(t) \) to find the maximum height.
\[
h(5) = -16(5)^2 + 160(5) = -16(25) + 800 = -400 + 800 = 400
\]

Step1: Set height to zero

To find when the ball reaches the ground, set \( h(t) = 0 \) and solve for \( t \).
\[
-16t^2 + 160t = 0
\]

Step2: Factor the equation

Factor out \( -16t \) from the left side.
\[
-16t(t - 10) = 0
\]

Step3: Solve for \( t \)

Set each factor equal to zero and solve for \( t \).

• \( -16t = 0 \) gives \( t = 0 \) (this is the initial time when the ball is kicked).
• \( t - 10 = 0 \) gives \( t = 10 \) (this is the time when the ball reaches the ground).

Answer:

\( h_{\text{max}} = \boxed{400} \) feet

Part (b)