Question

name________________________ date________________________
choice 3: the flight of a soccer ball
the function ( s(t)=-16 t^{2}+64 t ) represents the height in feet of a soccer ball t seconds after it was kicked by a player.

1. without graphing, find ( s(3) ). what does this value mean in terms of the soccer balls flight?
2. without graphing, determine the time it takes for the soccer ball to reach its maximum height. what was the maximum height?
3. without graphing, determine how long the soccer ball is in the air.
4. without graphing, determine the height from which the soccer ball is kicked.
5. what domain could be used to describe the soccer balls flight through the air? range?

Explanation:

Step1: Calculate \(s(3)\)

Substitute \(t = 3\) into \(s(t)=-16t^{2}+64t\).

$$
LATEXBLOCK0
$$

This means that 3 seconds after the soccer ball was kicked, its height is 48 feet.

Step2: Find the time to reach maximum height

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

$$ t=-\frac{64}{2\times(-16)}=\frac{-64}{-32}=2 $$

Substitute \(t = 2\) into \(s(t)\) to find the maximum height.

$$
LATEXBLOCK1
$$

Step3: Determine how long the ball is in the air

Set \(s(t)=0\), so \(-16t^{2}+64t=0\). Factor out \(-16t\): \(-16t(t - 4)=0\).
Using the zero - product property, \(-16t=0\) gives \(t = 0\) (initial time) and \(t-4=0\) gives \(t = 4\). So the ball is in the air for 4 seconds.

Step4: Find the height from which the ball is kicked

Set \(t = 0\) in \(s(t)\). \(s(0)=-16\times(0)^{2}+64\times0=0\).

Step5: Determine domain and range

The domain (time \(t\)): Since the ball is in the air from \(t = 0\) to \(t = 4\), the domain is \(0\leq t\leq4\).
The range (height \(s(t)\)): The minimum height is \(s(0)=s(4)=0\) and the maximum height is \(s(2) = 64\). So the range is \(0\leq s(t)\leq64\).

Answer:

1. \(s(3)=48\), meaning the ball's height is 48 feet 3 seconds after being kicked.
2. Time to reach maximum height: \(t = 2\) seconds, maximum height: \(64\) feet.
3. The ball is in the air for \(4\) seconds.
4. The ball is kicked from a height of \(0\) feet.
5. Domain: \(0\leq t\leq4\), Range: \(0\leq s(t)\leq64\).