Question

14. A volleyball is hit upward by a player in a game. The height h (in feet) of the volleyball after t (seconds) is given by h (t) = -5ť^2 + 30t + 6.

Answer 1

We have that the function that represents this situation is:

$h(t)=-5t^2+30t+6$

notice that we have a quadratic function, then, we can find the maximum height by finding the vertex of the parabolla. To find the vertex, we can use the following general rule:

$\begin{gathered} f(x)=ax^2+bx+c \\ vertex\colon(-\frac{b}{2a},f(-\frac{b}{2a})) \end{gathered}$

in this case, we have the following coefficients:

$a=-5,b=30,c=6$

then, the x coordinate of the vertex is:

$-\frac{b}{2a}=-\frac{30}{2(-5)}=-\frac{30}{-10}=3$

finally, we can evaluate f(3) to find the y-coordinate:

$\begin{gathered} f(-\frac{b}{2a})=f(3)=-5(3)^2+30(3)+6=-5(9)+90+6 \\ =-45+96=51 \end{gathered}$

we have that f(3) = 51. This means that at time t = 3 seconds, the volleyball reaches its maximum height of 51ft