. A volleyball is served from a height of 6 feet with an initial velocity of 33 feed per second. The situations modeled using the function , where is the height in feed and t is the time in seconds. Using the function, what is the vertex?

1 answer

To model the height of the volleyball over time, you can use the equation of a projectile motion. The general form of the height \( h \) as a function of time \( t \) can be given by the formula:

\[ h(t) = -16t^2 + vt + h_0 \]

where:

  • \( v \) is the initial velocity (in feet per second),
  • \( h_0 \) is the initial height (in feet),
  • and the coefficient \(-16\) represents the effect of gravity in feet per second squared.

Given:

  • Initial height \( h_0 = 6 \) feet
  • Initial velocity \( v = 33 \) feet per second

Substituting these values into the equation:

\[ h(t) = -16t^2 + 33t + 6 \]

To find the vertex of the parabola represented by this equation, we can use the formula for the vertex \( t = -\frac{b}{2a} \) where \( a \) and \( b \) are the coefficients from the standard form \( at^2 + bt + c \).

Here:

  • \( a = -16 \)
  • \( b = 33 \)

Now substituting these values:

\[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \text{ seconds} \]

Now we can find the height at this time:

\[ h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6 \]

Calculating \( \left(\frac{33}{32}\right)^2 \):

\[ \left(\frac{33}{32}\right)^2 = \frac{1089}{1024} \]

Substituting back into the height equation:

\[ h\left(\frac{33}{32}\right) = -16 \cdot \frac{1089}{1024} + 33 \cdot \frac{33}{32} + 6 \] \[ h\left(\frac{33}{32}\right) = -\frac{17424}{1024} + \frac{1089}{32} + 6 \] \[ = -\frac{17424}{1024} + \frac{34656}{1024} + \frac{6144}{1024} \] \[ = \frac{17424 - 17424 + 6144}{1024} \] \[ = \frac{6144}{1024} = 6 \]

As a result, the vertex of the function is:

\[ \left(\frac{33}{32}, 6\right) \]

In conclusion, the vertex of the function representing the height of the volleyball is at:

\[ \left(\frac{33}{32}, 6\right) \approx (1.03, 6) \text{ (in feet)} \]

This means the maximum height reached is 6 feet at approximately \( t = 1.03 \) seconds.