Question 23297
The equation is:
{{{s(t) = -16t^2 + 64t}}} To find the time (t) at which the ball returns to the ground (s=0), set the above equation equalto 0 and solve for t.
{{{0 = -16t^2+64t}}} Factor out a t.
{{{0 = t(-16t+64)}}} Apply the zero product principle.
{{{t = 0}}} and/or {{{-64t+64 = 0}}}
The solution t=0 would be the situation at the start of this operation. So we are left with:
{{{-16t+64 = 0}}} Add 64t to both sides of the equation.
{{{64 = 16t}}} Divide both sides by 16.
{{{t = 4}}}

The ball returns to the ground at t = 4 seconds.

To find the maximum height reached by the ball you first will find the time, t, at which the function s(t) is a maximum.  Recall that the quadratic function represents a parabola and the maximum (or minimum) point on a parabola occurs at its vertex. The vertex, in this case, will be a maximum value because the parabola opens downward. How do you know this?...the coefficient of t^2 is negative.

The x-coordinate (or, in this case, the t-coordinate) of the vertex is given by:{{{t = -b/2a}}} where a=-16 and b=64.  This is taken from the standard form for the quadratic equation: {{{at^2+bt+c=0}}}

The t-coordinate of the vertex is:
{{{t = (-64)/2(-16)}}}
{{{t = (-64)/(-32)}}}
{{{t = 2}}} The maximum height occurs at time t=2 seconds. Find the maximum height by substituting this value of t into the original function: {{{s(t) = -16t^2+64t}}}

{{{s(2) = -16(2)^2 + 64(2)}}}
{{{s(2) = -64+128}}}
{{{s(2) = 64}}}

The maximum height attained by the ball is 64 feet at 2 seconds.