Question 24543
Hmmm...I think I have answered this one previously. Anyway, here it is again.
The equation is correct:
{{{s(t) = -16t^2 +vot + ho}}} 
{{{s(t) = -16t^2 + 64t}}} This is the function describing the problem. In words, the height (s) is a quadratic function of time (t) with initial velocity of 64 ft/sec and an initial height of zero.

After 1 second (t=1), the height of the baseball will be:
{{{s(1) = -16(1)^2 + 64(1)}}}
{{{s(1) = -16+ 64}}}
{{{s(1) = 48}}} The height will be 48 feet after 1 second.

Find when the baseball will hit the ground (s=0) by setting the function s(t) = 0 and solving for t.
{{{-16t^2 + 64t = 0}}} Factor out a t.
{{{t(-16t + 64) = 0}}} Divide both sides by t.
{{{-16t + 64 = 0}}} Subtract 64 from both sides.
{{{-16t = -64}}} Divide both sides by -16
{{{t = 4}}} The baseball will return to ground in 4 seconds.

The maximum height and time is found by finding the location of the vertex of this parabola.
Since the parabola opens downward, the vertex will be at the maximum point of the vertex and will represent the maximum height attained by the baseball.
The x-coordinate (or, in this problem, the t-coordinate) is given by:
{{{t = -b/2a}}}
The a and b come from the standard form of the quadratic equation: {{{ax^2 + bx + c = 0}}}
In this case, a = -16, b = 64, and c = 0

{{{t = -64/2(-16)}}}
{{{t = 2}}} The maximum height of the baseball will be attained in 2 seconds.

To find this maximum eight, substitute t=2 into the original function and solve for s.
{{{S(2) = -16(2)^2 + 64(2)}}}
{{{s(2) = -16(4) + 128}}}
{{{s(2) = -64 + 128}}}
{{{s(2) = 64}}} The maximum height attained by the baseball is 64 feet.