## Trigonometry (10th Edition)

The balloon's speed will be 15.8 ft/s The balloon's path will make an angle of $71.6^{\circ}$ with the horizontal.
The hot-air balloon's resultant vector is $(5.00~ft/s, 15.0~ft/s)$ We can find the magnitude of the vector which is the balloon's speed: $v = \sqrt{(5.00~ft/s)^2+(15.0~ft/s)^2}$ $v = 15.8~ft/s$ The balloon's speed will be 15.8 ft/s We can find the angle above the horizontal: $tan~\theta = \frac{15.0}{5.00}$ $\theta = arctan(\frac{15.0}{5.00})$ $\theta = 71.6^{\circ}$ The balloon's path will make an angle of $71.6^{\circ}$ with the horizontal.