Question 204443
Starting with:
{{{h(d) = -0.09d^2+0.9d+2}}}
a) The maximum h can be found at the vertex of the parabola at:
{{{d = (-0.9)/2(-0.09)}}}
{{{d = 5}}} Now substitute this value of d into the given equation and solve for h.
{{{h(5) = 0.09(5)^2+0.9(5)+2}}}
{{{h(5) = 4.25}}}
The maximum height is 4.25 metres.
b) The distance d when the ball has reached a height of 4.25 metres is: 
{{{4.25 = -0.09d^2+0.9d+2}}} Subtract 4.25 from both sides.
{{{-.09d^2+0.9d-2.25 = 0}}} Solve using the quadratic formula:{{{x = (-b+-sqrt(b^2-4ac))/2a}}} where: a = -0.09, b = 0.9, and c = 2.25 and instead of x in the formula, we want to use d, so...
{{{d = (-0.9+-sqrt(0.9^2-4(-0.09)(2.25)))/2(-0.09)}}}
{{{d = (-0.9+-sqrt(0.81-(-0.81)))/(-0.18)}}}
{{{d = (-0.9+-sqrt(1.62))/(-0.18)}}}
{{{d = 5}}}metres.
c) From the original equation, the ball is 2 metres from the floor when the player releases it.
The general form of the equation for the height (h) an object propelled upward  with an initial velocity {{{v[0]}}} from an initial height of {{{h[0]}}} is:
{{{h(t) = -16t^2+v[0]t+h[0]}}} and {{{h[0]}}} = 2.