Question 149814
Find the maximum height, h, of the ball:
{{{h(t) = -16t^2+128t}}} This quadratic equation represents a downward-opening parabola.  The maximum value of the indepedent variable (that's h in this case) occurs at the vertex of the parabola. The t-coordinate of the vertex is found by:
{{{t = (-b)/2a}}} where a = -16 and b = 128, so...
{{{t = (-128)/2(-16)}}}
{{{t = (-128)/(-32)}}}
{{{t = 4}}} The maximum height is attained at t = 4 seconds.
To find the height at this time, substitute t = 4 into the given equation and solve for h.
{{{h(4) = -16(4)^2+128(4)}}}
{{{h(4) = -256+512}}}
{{{h(4) = 256}}}feet. This is the maximum height attained by the ball.