check the picture below, that one use feet, but is the same thing, in this case is just centimeters.
a)
so the parabola highest point is at the vertex's y-coordinate, which one will that be?
[tex]\bf h=-4t^2+60\implies h=-4t^2+0t+60
\\\\\\
\textit{vertex of a vertical parabola, using coefficients}
\\\\
\begin{array}{lcccl}
h = & -4t^2& +0t& +60\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad
\left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(~~~~~~,~~60-\cfrac{0^2}{4(-4)} \right)\implies \left(~~~~~~,~~60-0 \right)\implies (~~~~,~60)[/tex]
b)
well, when t = 3, you'll just get h = -4(3)² + 60.
and surely you know what that is.
c)
recall that the x-intercepts occur when y = 0, in this case, when h = 0,
[tex]\bf h=-4t^2+60\implies \stackrel{h}{0}=-4(t^2-15)\implies 0=t^2-15
\\\\\\
0=\stackrel{\textit{difference of squares}}{t^2-(\sqrt{15})^2}\implies 0=(t-\sqrt{15})(t+\sqrt{15})\\\\
-------------------------------\\\\
0=t-\sqrt{15}\implies \sqrt{15}=t\\\\
-------------------------------\\\\
0=t+\sqrt{15}\implies -\sqrt{15}=t[/tex]
