Question

question 14
0/1 pt 3 99 details
a cannonball is fired in the air at an angle of 45°. how far does it travel before it is 1,800 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.)
feet
question help: video

Explanation:

Step1: Use right - triangle trigonometry

Since the cannonball is fired at a 45 - degree angle and we assume a straight - line path, the vertical and horizontal components of the motion are equal. If the vertical height $y = 1800$ feet and we consider a right - triangle formed by the path of the cannonball, horizontal distance $x$, and the path of the cannonball itself. In a 45 - 45 - 90 right - triangle, $\tan\theta=\frac{y}{x}$. When $\theta = 45^{\circ}$, $\tan45^{\circ}=1=\frac{y}{x}$.

Step2: Solve for the horizontal distance

Since $\tan45^{\circ}=\frac{y}{x}=1$ and $y = 1800$ feet, then $x=y = 1800$ feet.

Answer:

1800