Solve the problem.
-A projectile is fired from a cliff 300 feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of 270 feet per second. The height h of the projectile above the water is given by $$h ( x ) = \frac { - 32 x ^ { 2 } } { ( 270 ) ^ { 2 } } + x + 300$$ ,
Where x is the horizontal distance of the projectile from the base of the cliff. Find the maximum height of the
Projectile.
A) 869.53 ft
B) 569.53 ft
C) 2,008.59 ft
D) 1,139.06 ft

Question

Quizplus · Accepted Answer

The maximum height of the projectile can be found by completing the square for the given quadratic equation. The given equation is in the form of $h(x) = ax^2 + bx + c$, where $a = \frac{-32}{(270)^2}$, $b = 1$, and $c = 300$. To find the vertex of the parabola, which gives the maximum height (since $a < 0$), we use the formula for the x-coordinate of the vertex, $x = -\frac{b}{2a}$. Plugging in the values, we get the x-coordinate of the vertex. Substituting this back into the original equation gives the maximum height, which is approximately 869.53 feet.

Solve the Problem h(x)=−32x2(270)2+x+300h ( x ) = \frac { - 32 x ^ { 2 } } { ( 270 ) ^ { 2 } } + x + 300h(x)=(270)2−32x2​+x+300

Solve the Problem $h ( x ) = \frac { - 32 x ^ { 2 } } { ( 270 ) ^ { 2 } } + x + 300$