The height in feet, y, of a baseball t seconds after it is hit is given by the following formula:

Question

                y = −16t2 + 88t

The height of the baseball at each half-second interval is shown in the table below.

Time (seconds)Height (feet)0.000.5401.0721.5962.01122.51203.01203.51124.0964.5725.0405.50
According to the formula, what was the greatest height, in feet, the baseball reached?

Answer 1

The height of the baseball as a function of time is given by the equation:

\[ y = -16t^2 + 88t \]

This is a quadratic equation that opens downward (since the coefficient of \( t^2 \) is negative). The greatest height (the vertex of the parabola) of the function can be found using the vertex formula. For a quadratic equation in standard form \( y = ax^2 + bx + c \), the time \( t \) at which the maximum height occurs is given by:

\[ t = -\frac{b}{2a} \]

In our equation, \( a = -16 \) and \( b = 88 \). Plugging in these values, we get:

\[ t = -\frac{88}{2 \times -16} = \frac{88}{32} = 2.75 \text{ seconds} \]

Now we substitute \( t = 2.75 \) back into the height equation to find the maximum height:

\[ y = -16(2.75)^2 + 88(2.75) \]

Calculating \( (2.75)^2 \):

\[ (2.75)^2 = 7.5625 \]

Now substituting into the equation gives:

\[ y = -16(7.5625) + 88(2.75) \]

Calculating \( -16(7.5625) \):

\[ -16(7.5625) = -121 \]

Calculating \( 88(2.75) \):

\[ 88(2.75) = 242 \]

Now, combining both parts:

\[ y = -121 + 242 = 121 \]

Thus, the greatest height the baseball reached is:

\[ \boxed{121 \text{ feet}} \]