Chapter 11.4, Problem 42E

Path of a Cannonball 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 3200 ft above the ground, find an equation for the path of the cannonball. Place the origin at the location of the cannon.

To determine

To find: The equation of the path of the cannon ball.

Answer to Problem 42E

The equation of the path of the cannon ball is ${\left(x-800\right)}^2=-200\left(y-3200\right)$.

Explanation of Solution

Given:

The cannon ball lands 1600ft from the cannon and the highest point it reaches is 3200ft and the cannon located at the origin.

Definition used:

Definition 1:

“The equation of shifted parabola with vertex $V\left(h,k\right)$ and the axis parallel to x-axis is ${\left(x-h\right)}^2=4p\left(y-k\right)$”.

Definition 2:

“The equation of the parabola vertex $V\left(0,0\right)$, focus $F\left(0,p\right)$ and directrix $y=-p$ is $x^2=4py$.”

Definition 3:

“If $h$ and $k$ are any two positive real numbers and a graph of any equation is in terms of $x$ and $y$ then.

(i) Replacing $x$ by $x-h$, shift the graph right to $h$ units.

(ii) Replacing $x$ by $x+h$, shift the graph left to $h$ units.

(iii) Replacing $y$ by $y-k$, shift the graph upward to $k$ units.

(iv) Replacing $y$ by $y+k$, shift the graph downward to $k$ units.”

Calculation:

The given canon is placed at the origin, if the ball is fired the path of the ball is parabolic and the highest point is the vertex of the parabola.

Let the coordinate axes of the parabola meet at O as shown in Figure 1.

Also, the ball strikes the ground at 1600ft from the origin $O\left(0,0\right)$.

Therefore, the coordinate of the point where the ball strikes the ground is $A\left(1600,0\right)$.

The midpoint of OA is $\left(800,0\right)$ and the line perpendicular to the x-axis.

That is, $x=800$ is the axis of the parabola.

The highest point reached by the ball is 3200ft above the ground.

Therefore, vertex of the parabolic path is $V\left(800,3200\right)$ as shown in Figure 1.

By the definition 1, the equation of the path of the cannon ball is ${\left(x-800\right)}^2=4p\left(y-3200\right)$ (1)

From Figure 1, it is clear that the equation (1) passes through $\left(0,0\right)$.

$\begin{array}{l}{\left(0-800\right)}^2=4p\left(-3200\right)\\ 640000=-12800p\\ p=-\frac{640000}{12800}\\ p=-50\end{array}$

Substitute $p=-50$ in equation (1),

$\begin{array}{l}{\left(x-800\right)}^2=4\left(-50\right)\left(y-3200\right)\\ {\left(x-800\right)}^2=-200\left(y-3200\right)\end{array}$

Thus, the equation of the path of the cannon ball is ${\left(x-800\right)}^2=-200\left(y-3200\right)$.