Question 1011428
A projectile is fired at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 100 feet per second. 
The height H of the projectile is modeled by 
{{{ H(x) = -32x^2/(100)^2+x}}}
Where x is the horizontal distance of the projectile from the firing point.

(A) At what horizontal distance from the firing point is the height of the
projectile a maximum?
This is quadratic equation, the max will occur at the axis of symmetry; x=-b(2a)
Find a: -32/100^2 = .0064, b = 1
x = {{{(-1)/(2*-.0032)}}}
x = +156.25 ft
:
(B) Find the maximum height of the projectile.
x = 156.25
{{{H(x) = -32(156.25^2)/(100)^2+156.25}}}
H(x) = 78.125 ft is the max ht 
:
(C) At what horizontal distance from the firing point will the projectile strike the ground?
It will be twice the axis of symmetry
x = 2(156.25)
x = 312.5 ft from the origin when it hits the ground
:
(D) Using a graphing utility, graph the function H,
 0(is equal to or less than) x (is equal to or less than) 350
Should look like this
{{{ graph( 300, 200, -100, 350, -10, 100, (-32x^2/100^2)+x, 50) }}}


(E) Using the graphing utility to verify the results obtained in parts B and C

(F) when the height of the projectile is 50 feet above the ground, how far has it traveled horizontally?
h = 50
{{{(-32x^2/100^2)+x}}}= 50

{{{-.00032x^2 + x - 50 = 0}}}
Solve this with the quadratic formula; a=-.0032, b=1, c=-50
Two solutions
x = 62.5 ft on the way up
and
x = 250 ft on the way down
(Green line on the graph is 50')