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 300 feet per second. The height h of the projectile above the water is given by $$h ( x ) = \frac { - 32 x ^ { 2 } } { ( 300 ) ^ { 2 } } + x + 300$$ 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) 1,003.13 ft
B) 1,406.25 ft
C) 703.13 ft
D) 2,409.38 ft

Question

Quizplus · Accepted Answer

The maximum height of a projectile in a parabolic trajectory occurs at the vertex of the parabola. The x-coordinate of the vertex for a quadratic equation $ax^2 + bx + c$ is given by $-b/(2a)$. Here, $a = -32/(300)^2$ and $b = 1$. Calculating $-b/(2a)$ gives approximately 1,406.25 ft.

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

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