Question 169248
You do have a small but significant error in your equation. The general form for the function of the height (as a function of time) of an object propelled upward at an initial velocity of {{{v[0]}}} from an initial height of {{{h[0]}}} is given by:
{{{h(t) = -16t^2+v[0]t+h[0]}}} so your equation should read:
{{{h(t) = -16t^2+48t}}} and, for your problem, {{{v[0] = 48}}} and {{{h[0] = 0}}}.
You want to find the maximum height of the ball which means you need to know the location of the vertex (a maximimum in this case) of the parabola represented by the equation.
The t-coordinate of this parabola is given by:
{{{t = -b/2a}}} This will give you the time at which the ball reaches the maximum height. Once you have this, you can then find the maximum height.
{{{t = -(48)/2(-16)}}}
{{{t = 3/2}}}seconds.  Now substitute this value of t into the original equation and solve for the height, h.
{{{h(3/2) = -16(3/2)^2+48(3/2)}}}
{{{h(3/2) = -(16(9/4)) + 72}}}
{{{h(3/2) = -36+72}}}
{{{highlight(h(3/2) = 36)}}}feet. This is the maximum height reached by the ball.
{{{graph(400,400,-5,5,-5,40,-16x^2+48x)}}}