Question 31632
Using the given function:
{{{s(t) = -16t^2 + (vo)t + so}}} and substituting vo = 64 ft/sec. and so = 0
{{{s(t) = -16t^2 + 64t}}} Set s(t) = 0 and solve for t to find the time at which the baseball will return to the ground.
{{{0 = -16t^2 + 64t}}} Solve for t. First factor out a t.
{{{0 = t(-16t + 64)}}} Apply the zero products principle.
{{{t = 0}}} This is the initial condition when the height, s = 0.
{{{0 = -16t + 64}}} Add 16t to both sides of the equation.
{{{16t = 64}}} Divide both sides by 16.
{{{t = 4}}}Secs.
1) The ball returns to the ground in 4 seconds.

The maximum height of the ball can be found by finding the location of the verttex of the parabola that is represented by the original quadratic equation{{{s(t) = -16t^2 + 64t}}}
The t-coordinate of the vertex is given by{{{t = (-b)/2a}}}
The a and b come from the standard form of a quadratic equation{{{ax^2+bx+c=0}}}
In this problem, a = -16, b = 64, and c = 0

{{{t = (-64)/2(-16)}}}
{{{t = (-64)/-32}}}
{{{t = 2}}}Secs. This is the time at which the baseball reaches its maximum height.
To find the maximum height, substitute this value of t into the original equation and solve for s.

{{{s(2) = -16(2)^2 + 64(2)}}}
{{{s(2) = -64 + 128}}}
{{{s(2) = 64}}}Feet.  This is the maximum height attained by the baseball.