Answer:
A) 0.2 seconds
Step-by-step explanation:
[tex]b(t)=-16t^2+5t+15[/tex]
[tex]v(t)=-32t+5[/tex] <-- Take the derivative
[tex]0=-32t+5[/tex] <-- Set equal to 0
[tex]-5=-32t[/tex]
[tex]\frac{5}{32}=t[/tex]
[tex]t=\frac{5}{32}[/tex]
[tex]t\approx0.2[/tex]
Therefore, it will take the ball about 0.2 seconds to reach its maximum height (which happens to be about 15.4 feet BTW).
Alternatively, another non-calculus approach is the fact that the parabola opens downward, and as such, the maximum of [tex]b(t)[/tex] occurs at the vertex where [tex]t=-\frac{b}{2a}[/tex], therefore [tex]t=-\frac{5}{2(-16)}=\frac{-5}{-32}=\frac{5}{32}\approx0.2[/tex]
