Question 10673
{{{y = 400x - 16^2}}}

This is the equation of the height of an object propelled upward at an intial velocity of 400 ft/sec at an initial height of zero ft. It is usually written as:

{{{h = -16t^2 + 400t}}} Where: h = height and t = time.

The maximum height occurs at the vertex of the parabola whose x-coordinate (or t-value) is given by: {{{-b/2a}}}.  This value of t will be the time at which the object reaches its maximum height. The equation must be in the standard form of:
{{{ax^2 + bx + c}}} or, in this case: {{{at^2 + bt + c}}} But here, a = -16, b = 400, and c = 0.

So, {{{(-b/2a) = (-400/(2(-16)))}}} = 12.5 secs. The time at maximum height.

Now, we substitute this value of t into the original eqation and solve for h to get the maximum height.

{{{-16(12.5)^2 + 400(12.5) = -2500 + 5000}}} = 2500 feet. This is the maximum height attained by the object.