Chapter 3, Problem 3T

A cannonball fired out to sea from a shore battery follows a parabolic trajectory given by the graph of the equation

$h(x)=10x-0.01x^2$

where h(x) is the height of the cannonball above the water when it has traveled a horizontal distance of x feet.

1. (a) What is the maximum height that the cannonball reaches?
2. (b) How far does the cannonball travel horizontally before splashing into the water?

To determine

To find: Maximum height of the cannonball.

Answer to Problem 3T

The maximum height of the cannonball that travels is $2500\text{feet}$.

Explanation of Solution

Definition used:

The parabolic equation of degree 2 is defined as,

$a{x}^2+bx+c$ (1)

Given:

The parabolic trajectory path in which cannonball travels is,

$h\left(x\right)=10x-0.01x^2$. (2)

Calculation:

Compare equation (1) and (2) for the values of the constant,

Therefore, $a=-0.01$ and $b=10$

As the value of a is negative, the graph will have its head up and tail down so that the ratio of the maximum point to the minimum point is a maximum point.

The value of the maximum point is given by the expression $-\frac{b}{2a}$,

Substitute 10 for $b$ and $-0.01$ for $a$ in the expression $-\frac{b}{2a}$ for the maximum point,

$\begin{array}{c}\frac{-b}{2a}=-\frac{\left(10\right)}{2\times \left(-0.01\right)}\\ =500\end{array}$

Therefore the value of the maximum point is at $500$.

Substitute 500 for x in equation (2) for the maximum point of the height.

$\begin{array}{c}h\left(x\right)=10x-0.01x^2\\ h\left(500\right)=10\times 500-0.01{\left(500\right)}^2\\ =2500\end{array}$

Therefore, maximum height is $2500$ feet when the cannon ball has reached a horizontal distance of $500$ feet.

To determine

To find: Distance travelled by cannonball in horizontal direction.

Answer to Problem 3T

The distance travelled by cannonball in horizontal direction before splashing in water is $1000\text{feet}$.

Explanation of Solution

Given:

The parabolic trajectory path in which cannonball travels is,

$h\left(x\right)=10x-0.01x^2$. (1)

Calculation:

Substitute $0$ for $h\left(x\right)$ in equation (1) for the maximum horizontal distance that the cannonball travels before splashing in water,

$\begin{array}{c}10x-0.01x^2=0\\ x\left(10-0.01x\right)=0\end{array}$

Therefore, solve the above expression,

$x=0$ and $10-0.01x=0$

Solve further the above equation $10-0.01x=0$.

$\begin{array}{c}10-0.01x=0\\ x=1000\end{array}$

Thus, the maximum distance travelled by cannonball in horizontal direction before splashing in water is $1000\text{feet}$.