Step-by-step explanation:
The height of the ball is given by
[tex]h = -16t^2 + 144t + 5[/tex]
We can solve for the time when it reaches the height of 325 feet as follows:
[tex]325 = -16t^2 + 144t + 5 \\ \Rightarrow 16t^2 - 144t + 320=0[/tex]
To solve this equation, we use the quadratic formula:
[tex]t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2s}[/tex]
[tex]\:\:\:\:= \dfrac{144 \pm \sqrt{(-144)^2 - 4(16)(320)}}{32}[/tex]
[tex]\:\:\:\:=\dfrac{144 \pm 16}{32} = 4\:\text{s},\:5\:\text{s}[/tex]
So this means that the ball will reach the height of 325 ft four seconds after launch, continue going up and a second later, it will at the height of 325 ft on it way down.
