A cannon fires a cannonball as shown in the figure. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannonball lands 1600 ft from the cannon and the highest point it reaches is 2400 ft above the ground, find an equation for the path of the cannonball. Place the origin at the location of the cannon.
A) $( x - 1200 ) ^ { 2 } = - \frac { 400 } { 3 } ( y - 800 )$
B) $( x - 800 ) ^ { 2 } = - \frac { 400 } { 3 } ( y - 2400 )$
C) $( x - 800 ) ^ { 2 } = - \frac { 800 } { 3 } ( y - 2400 )$
D) $( x - 2400 ) ^ { 2 } = - \frac { 800 } { 3 } ( y - 800 )$
E) none of these

Question

A cannon fires a cannonball as shown in the figure. The path of the cannonball is a parabola with vertex at the highest point of the path. If the cannonball lands 1600 ft from the cannon and the highest point it reaches is 2400 ft above the ground, find an equation for the path of the cannonball. Place the origin at the location of the cannon. 
A) $( x - 1200 ) ^ { 2 } = - \frac { 400 } { 3 } ( y - 800 )$ 
B) $( x - 800 ) ^ { 2 } = - \frac { 400 } { 3 } ( y - 2400 )$ 
C) $( x - 800 ) ^ { 2 } = - \frac { 800 } { 3 } ( y - 2400 )$ 
D) $( x - 2400 ) ^ { 2 } = - \frac { 800 } { 3 } ( y - 800 )$ 
E) none of these

Quizplus · Accepted Answer

We first note that the path of the cannonball is a parabola with a vertical axis of symmetry, since it goes up and then down. This means the equation for the parabola will be of the form $(x-h)^2=a(y-k)$, where (h,k) is the vertex.

Since the highest point is 2400 ft above the ground, the y-coordinate of the vertex is 2400. Since the cannonball lands 1600 ft from the cannon, the x-coordinate of the vertex is 800. We can plug these values into the vertex form to get $(x-800)^2=a(y-2400)$.

Now we just need to find the value of $a$. We know that the vertex is the highest point of the parabola, so we can use this to find $a$. At the vertex, the derivative of the equation with respect to y is 0. Taking the derivative and setting it equal to 0 gives us $\frac{dx}{dy}=a$, so the value of $a$ is just the horizontal distance the cannonball travels divided by four times the height it reaches: $a=\frac{1600}{4	imes 2400/3}= \frac{1}{3}$.

Plugging this value of $a$ back into the vertex form gives us $(x-800)^2=\frac{-800}{3}(y-2400)$. Multiplying both sides by $-3/800$ gives us $\boxed{(C)\ (x-800)^2=-\frac{800}{3}(y-2400)}$ as the equation of the path of the cannonball.

A Cannon Fires a Cannonball as Shown in the Figure (x−1200)2=−4003(y−800)( x - 1200 ) ^ { 2 } = - \frac { 400 } { 3 } ( y - 800 )(x−1200)2=−3400​(y−800)

A Cannon Fires a Cannonball as Shown in the Figure $( x - 1200 ) ^ { 2 } = - \frac { 400 } { 3 } ( y - 800 )$