A projectile is fired from a cliff 500 feet above the water

A projectile is fired from a cliff 500 feet above the water at an inclination of 45° to the horizontal, with a muzzle
velocity of 170 feet per second. The height h of the projectile above the water is given by h(x) =
-32x2/
(170)2 + x + 500,
where x is the horizontal distance of the projectile from the base of the cliff. How far from the base of the cliff is
the height of the projectile a maximum?

a. 725.78 ft
b. 225.78 ft
c. 1177.34 ft
d. 451.56 ft

Solution

h(x) = -32x^2/(170)^2 + x + 500

The x-coordinate of the vertex of any parabola in the form f(x) = ax^2 + bx + c is given by -b/2a .

So evaluate -1/[2*-32/(170)^2] = 451.56 m  to find the value of x, i.e. the horizontal distance, at the vertex, or maximum