To solve this problem, let's go through the steps methodically.
1. Expression for the height of the ball:
The height of the ball as a function of time [tex]\( t \)[/tex] is given by the quadratic equation:
[tex]\[ h = -8t^2 + 40t \][/tex]
2. Find the zeros of the equation:
To find the time at which the ball touches the ground (height [tex]\( h \)[/tex] becomes zero), we need to solve the equation:
[tex]\[ -8t^2 + 40t = 0 \][/tex]
We can factor out [tex]\( -8t \)[/tex] from the equation:
[tex]\[ -8t(t - 5) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \quad \text{or} \quad t = 5 \][/tex]
These are the times when the ball is at the ground level.
3. Determine the time at which the ball reaches its highest point:
Since the vertex of a parabola is symmetric and lies exactly halfway between the zeros, we calculate the midpoint between [tex]\( t = 0 \)[/tex] and [tex]\( t = 5 \)[/tex]:
[tex]\[ t_{\text{highest}} = \frac{0 + 5}{2} = 2.5 \][/tex]
So, the ball reaches its highest point at [tex]\( t = 2.5 \)[/tex] seconds.
4. Calculate the highest point (height) of the ball:
We substitute [tex]\( t = 2.5 \)[/tex] back into the original height equation to find the highest height:
[tex]\[ h_{\text{highest}} = -8(2.5)^2 + 40(2.5) \][/tex]
Evaluating this expression:
[tex]\[ h_{\text{highest}} = -8 \times 6.25 + 40 \times 2.5 \][/tex]
[tex]\[ h_{\text{highest}} = -50 + 100 \][/tex]
[tex]\[ h_{\text{highest}} = 50 \][/tex]
So, the maximum height [tex]\( h \)[/tex] that the ball reaches is 50 units.
5. The ordered pair representing the highest point:
The highest point the ball reaches can be represented as the ordered pair [tex]\(( t_{\text{highest}}, h_{\text{highest}} )\)[/tex]:
[tex]\[ \left( 2.5, 50.0 \right) \][/tex]
Summary:
- The ball takes 2.5 seconds to reach its highest point.
- The ordered pair representing the highest point that the ball reaches is [tex]\( \left( 2.5, 50.0 \right) \)[/tex].
These results indicate the vertex of the parabola formed by the ball's trajectory.
