Question 61942
The function that relates the height, h, (in feet) of an object propelled upwards as a function of time, t, (in seconds) is given by:
{{{h(t) = -16t^2 + Vot + Ho}}} where: Vo is the initial upward velocity and Ho is the initial height of the object.

When graphed, the given quadratic equation will yield a parabola that opens downward. So to answer the question, you will need to find the maximum value of the function s(x).
The maximum value is at the vertex of the parabola and this is given by {{{-b/2a}}}. The a and b come from the general form of the quadratic equation: {{{f(x) = ax^2 + bx + c}}}.
In this problem, a = -16 and b= 200, so the x-coordinate of the vertex is:
{{{-200/2(-16) = 6.25}}} 

The maximum height reached by the firework is attained at time, t = 6.25 seconds and that is when the explosion should occur.
To find the height at t = 6.25 seconds, substitute t = 6.25 into the quadratic equation and solve for s.
{{{h(6.25) = -16(6.25)^2 + 200(6.25) + 40}}}
{{{h(6.25) = -625 + 1250 + 4}}}
{{{h(6.25) = 629}}}feet.