Question

a cannonball is fired in the air at an angle of 45°. how far does it travel before it is 500 feet above ground? (assume the cannonball travels in a straight line. ignore the force of gravity and wind resistance. round your answer to the nearest foot.)

Explanation:

Step1: Set up a right - triangle relationship

Since the cannonball is fired at a 45 - degree angle and we know the vertical height \(y = 500\) feet. In a right - triangle formed by the path of the cannonball, the vertical height \(y\), the horizontal distance \(x\), and the path of the cannonball (the hypotenuse), and because the angle of elevation \(\theta=45^{\circ}\), we know that \(\tan\theta=\frac{y}{x}\). For \(\theta = 45^{\circ}\), \(\tan45^{\circ}=1\), so \(x = y\). Also, using the sine function \(\sin\theta=\frac{y}{d}\), where \(d\) is the distance the cannonball travels (the hypotenuse of the right - triangle). Given \(\theta = 45^{\circ}\) and \(y = 500\) feet, and \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\).

Step2: Solve for the distance \(d\)

We know that \(\sin45^{\circ}=\frac{y}{d}\), so \(d=\frac{y}{\sin45^{\circ}}\). Substituting \(y = 500\) feet and \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), we get \(d=\frac{500}{\frac{\sqrt{2}}{2}}=500\sqrt{2}\approx500\times1.414 = 707\) feet.

Answer:

707