Solve the problem.
-A projectile is fired from a cliff 600 feet above the water at an inclination of $45 ^ { \circ }$ to the horizontal, with a muzzle velocity of 120 feet per second. The height $h$ of the projectile above the water is given by $h ( x ) = \frac { - 32 x ^ { 2 } } { ( 120 ) ^ { 2 } } + x + 600$, 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)712.5 ft
B)225 ft
C)937.5 ft
D)112.5 ft

Question

Quizplus · Accepted Answer

The maximum height of the projectile occurs at the vertex of the parabola described by the quadratic function. The vertex $x$-coordinate for $h(x) = ax^2 + bx + c$ is given by $-\frac{b}{2a}$. Here, $a = -\frac{32}{14400}$ and $b = 1$. Calculating $-\frac{b}{2a}$ gives 225 ft.

Solve the Problem 45∘45 ^ { \circ }45∘ To the Horizontal, with a Muzzle Velocity of 120 Feet

Solve the Problem $45 ^ { \circ }$ To the Horizontal, with a Muzzle Velocity of 120 Feet