Question

if a ball is thrown upward at 39.2 meters per second from the top of a building that is 25 meters high, the height of the ball can be modeled by ( s = 25 + 39.2t - 4.9t^{2} ), where ( t ) is the number of seconds after the ball is thrown. answer parts a through c. b. explain the meaning of the coordinates of the vertex for this model. choose the correct explanation below. a. the ball reaches its maximum speed of 98.2 meters per second in 5 seconds. b. the ball reaches its maximum height of 103.4 meters in 4 seconds. c. the ball hits the ground after 5 seconds at a speed of 98.2 meters per second. d. the ball reaches its maximum height of 4 meters in 103.4 seconds. c. over what time interval is the function increasing? what does this mean in relation to the ball? a. until ( t = 8 ) seconds; at 8 seconds, the ball reaches its maximum height and then falls. b. until ( t = 103.4 ) seconds; at 103.4 seconds, the ball reaches its maximum height and then falls. c. until ( t = 4 ) seconds; at 4 seconds, the ball reaches its maximum height and then falls. d. until ( t = 5 ) seconds; at 5 seconds, the ball reaches its maximum height and then falls.

Explanation:

Step1: Analyze the quadratic function \(S = 25+39.2t - 4.9t^{2}\)

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-4.9\), \(b = 39.2\), \(c = 25\)), the \(t\) - coordinate of the vertex is given by \(t=-\frac{b}{2a}\).
Substitute \(a=-4.9\) and \(b = 39.2\) into the formula \(t=-\frac{b}{2a}\):
\(t=-\frac{39.2}{2\times(-4.9)}=\frac{39.2}{9.8} = 4\)

Step2: Find the \(S\) - coordinate of the vertex

Substitute \(t = 4\) into the function \(S = 25+39.2t-4.9t^{2}\)
\(S=25+39.2\times4-4.9\times4^{2}\)
\(S=25 + 156.8-4.9\times16\)
\(S=25+156.8 - 78.4\)
\(S=103.4\)

Step3: Analyze the increasing - decreasing nature of the quadratic function

Since \(a=-4.9<0\), the parabola opens downwards. The function \(y = ax^{2}+bx + c\) is increasing on the interval \((-\infty,-\frac{b}{2a})\)

Answer:

For part b: B. The ball reaches its maximum height of 103.4 meters in 4 seconds.
For part c: C. Until \(t = 4\) seconds; at 4 seconds, the ball reaches its maximum height and then falls.