Question 48257
One approach to the solution to this problem is to recognise that that path described by the object is that of a parabbola which opens downward.  The maximum height reached by the object will be at the vertex of the parabola.
So if you can find the value of h at the vertex, you will have found the maximum height attained by the object, right?
Let's rewrite the equation in function form as:
{{{h(t) = -16t^2+32t}}} Compare this with the standard form for a quadratic equation:
{{{y = ax^2+bx+c}}}so in your equation, a = -16, b = 32, and c = 0

The x-coordinate (or t-coordinate in your problem) of the vertex is given by: {{{x = (-b)/2a}}}, in your problem, this would be:
{{{t = (-32)/(2(-16))}}}
{{{t = 1}}}second.