Question 66214
Given:
{{{s = -16t^2+Vot+Ho}}} and Vo = 32 ft/sec and Ho = 0
1) Write the function for this problem.  This can be done by writing the above equation in function form and substituting the given initial conditions for V0 and Ho.
{{{s(t) = -16t^2+32t}}}

2) The height of the ball after 1 second can be found by substituting t = 1 into the function s(t) and solving for s.
{{{s(1) = -16(1)^2+32(1)}}}
{{{s(1) = -16+32}}}
s(1) = 16}}}
The height after 1 second will be 16 feet above the ground.

3) To find out how long it will take to hit the ground (s = 0), set the function s(t) = 0 and solve for t.
{{{s(t) = -16t^2+32t}}} 
{{{0 = -16t^2+32t}}} Simplify and solve for t. Factor a t.
{{{0 = t(-16t+32)}}} Apply the zero product principle:
{{{t = 0}}} and {{{-16t + 32 = 0}}}
{{{t = 0}}} is one solution and this represents the initial condition when the ball is first thrown.
{{{-16t+32 = 0}}} Subtract 32 from both sides.
{{{-16t = -32}}} Divide both sides by -16.
{{{t = 2}}} is the other solution and this represents the terminal condition when the ball hits the ground after falling.

4) The maximum height of the ball occurs at the vertex of the curve (a parabola) represented by the function: {{{s(t) = -16t^2+32t}}}
The vertex of this parabola is a maximum because the curve opens downward as indicated by the negative coefficient of the t^2 term.
The t-coordinate of the vertex is given by:
{{{t = (-b)/2a}}} The a and b come from: {{{ax^2+bx+c = 0}}} the general form for the quadratic equation. In this case, a = -16 and b = 32, so:
{{{t = (-32)/2(-16)}}}
{{{t = 1}}} The maximum height occurs at time t = 1 second. To find the height in feet, substitute t = 1 into the function s(t) and solve for s.
{{{s(1) = -16(1)^2 + 32(1)}}} Simplify.
{{{s(1) = -16+32}}}
{{{s(1) = 16}}} The maximum height in feet.
Here's a graph of the function:
{{{graph(300,200,-5,5,-5,20,-16x^2+32x)}}}