If a ball is thrown upward at 19.6 meters per second from the top of a building that is 25 meters​ high, the height of the ball can be modeled by Upper S equals 25 plus 19.6 t minus 4.9 t squared​, where t is the number of seconds after the ball is thrown.

Question

Answer parts a through c.
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Part 1
a. Find the​ t-coordinate and​ S-coordinate of the vertex of the graph of this quadratic function.
The​ t-coordinate of the vertex of the graph of this quadratic function is tequals
  
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Answer 1

To find the vertex of the quadratic function \( S(t) = 25 + 19.6t - 4.9t^2 \), we can use the formula for the \( t \)-coordinate of the vertex of a parabola given by the equation \( ax^2 + bx + c \), which is

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

Here, \( a = -4.9 \) and \( b = 19.6 \).

Now, substituting the values into the formula:

\[ t = -\frac{19.6}{2 \cdot -4.9} \]

Calculating the denominator first:

\[ 2 \cdot -4.9 = -9.8 \]

Now substituting this value back into the equation for \( t \):

\[ t = -\frac{19.6}{-9.8} = \frac{19.6}{9.8} = 2 \]

So, the \( t \)-coordinate of the vertex is \( t = 2 \) seconds.

Next, we need to find the \( S \)-coordinate of the vertex by substituting \( t = 2 \) back into the original function \( S(t) \):

\[ S(2) = 25 + 19.6(2) - 4.9(2^2) \]

Calculating each term:

1. \( 19.6 \times 2 = 39.2 \)
2. \( 2^2 = 4 \)
3. \( 4.9 \times 4 = 19.6 \)

Putting it all together:

\[ S(2) = 25 + 39.2 - 19.6 \]

Performing the addition and subtraction:

1. \( 25 + 39.2 = 64.2 \)
2. \( 64.2 - 19.6 = 44.6 \)

Thus, the \( S \)-coordinate of the vertex is \( S = 44.6 \).

So, the final answers are:

• The \( t \)-coordinate of the vertex is \( t = 2 \).
• The \( S \)-coordinate of the vertex is \( S = 44.6 \).