Question 202853
The height (h) of an object propelled upwards as a function of time (t) is given by:
{{{h(t) = -(1/2)gt^2+v[0]t+h[0]}}} where g, the constant of acceleration due to gravity is 32ft/sec^2, {{{v[0]}}} is the initial upwards velocity, and {{{h[0]}}} is the initial height of the object.
In this problem, {{{v[0] = 64}}}ft/sec. and {{{h[0] = 6}}}ft.
Making the appropriate substitutions into the function above, we get:
{{{h(t) = -16t^2+64t+6}}}
This equation, when graphed, is a parabola that opens downwards, so we are looking for the maximum point (the vertex) on the curve which will give us the maximum height attained by the juggler's ball.
The value of the independent variable (t in this case) at the vertex is given by:
{{{t = (-b)/2a}}} where b = 64 and a = -16.
{{{t = (-64)/2(-16)}}}
{{{t = 2}}}seconds. This is the time, t, at which the juggler's ball reaches its maximum height.  To find the actual maximum height, we substitute t = 2 into the function above and solve for h.
{{{h(2) = -16(2)^2+64(2)+6}}} Evaluate.
{{{h(2) = -16(4)+128+6}}}
{{{h(2) = -64+128+6}}}
{{{h(2) = 70}}}
The maximum height reached by the ball is 70 feet.
{{{graph(400,400,-5,5,-5,74,-16x^2+64x+6)}}}