Chapter 5.5, Problem 1TU

TEST FOR UNDERSTANDING FOR EXAMPLE $5.14$

Find a formula for the path of the cannonball if the slope of the angle of elevation is $s=0.7$ and the initial velocity is $v_0=300$ feet per second. What is the maximum height the cannonball reaches? How far downrange does the maximum height occur?

EXAMPLE $5.14$FLIGHT OF A CANNONBALL

A cannonball fired from a cannon will follow the path of the parabola that is the graph of $y=y\left(x\right)$, where $y=-16\left(1+s^2\right){\left(x/v_0\right)}^2+sx$ feet. Here $s$ is the slope of inclination of the cannon barrel, $v_0$ is the initial velocity in feet per second, and the variable $x$ is the distance downrange in feet. Suppose that a cannon is elevated with a slope of $0.5$ (corresponding to an angle of inclination of about $27$ degrees) and the cannonball is given an initial velocity of $250$ feet per second. (See Figure $5.62$).