Answer:
The maximum height of the ball is: 14.0625ft
Step-by-step explanation:
The missing information is:
[tex]h(t) = 30t - 16t^2[/tex]
Required
The maximum height the ball attained
First, we calculate the time to reach the maximum height.
For a function: [tex]h(t) = at^2 + bt + c[/tex]
The maximum is: [tex]t = \frac{-b}{2a}[/tex]
So, the maximum of [tex]h(t) = 30t - 16t^2[/tex] is:
[tex]t = \frac{-b}{2a}[/tex]
Where
[tex]a = -16; b = 30[/tex]
So:
[tex]t = -\frac{30}{2*-16}[/tex]
[tex]t = -\frac{30}{-32}[/tex]
[tex]t = \frac{30}{32}[/tex]
[tex]t = 0.9375[/tex]
The maximum height is then calculated as:
[tex]h(t) = 30t - 16t^2[/tex]
[tex]h(0.9375) = 30 * 0.9375- 16 * 0.9375^2[/tex]
[tex]h(0.9375) = 28.125- 14.0625[/tex]
[tex]h(0.9375) = 14.0625[/tex]
