1. **Problem statement:** A baseball is hit with an initial velocity $v_0=27.0$ m/s at an angle $\theta=30^\circ$ above the horizontal. Find: a) The time the ball is in the air. b) The horizontal distance traveled before hitting the ground. c) The maximum height reached. 2. **Formulas and rules:** - Vertical and horizontal components of velocity: $$v_{0x} = v_0 \cos \theta$$ $$v_{0y} = v_0 \sin \theta$$ - Time of flight for projectile launched and landing at same height: $$t = \frac{2 v_{0y}}{g}$$ where $g = 9.8$ m/s$^2$ is acceleration due to gravity. - Horizontal range: $$R = v_{0x} \times t$$ - Maximum height: $$H = \frac{v_{0y}^2}{2g}$$ 3. **Calculations:** - Calculate components: $$v_{0x} = 27.0 \times \cos 30^\circ = 27.0 \times 0.866 = 23.38 \text{ m/s}$$ $$v_{0y} = 27.0 \times \sin 30^\circ = 27.0 \times 0.5 = 13.5 \text{ m/s}$$ a) Time in air: $$t = \frac{2 \times 13.5}{9.8} = \frac{27.0}{9.8} = 2.76 \text{ s}$$ b) Horizontal distance: $$R = 23.38 \times 2.76 = 64.53 \text{ m}$$ c) Maximum height: $$H = \frac{(13.5)^2}{2 \times 9.8} = \frac{182.25}{19.6} = 9.30 \text{ m}$$ 4. **Final answers (rounded to nearest hundredths):** a) Time in air = 2.76 s b) Distance traveled = 64.53 m c) Maximum height = 9.30 m These results show the ball stays in the air for about 2.76 seconds, travels horizontally about 64.53 meters, and reaches a peak height of 9.30 meters.