## Physics: Principles with Applications (7th Edition)

Let x= 0 and y = 0 directly below where the baseball is launched, and let upward be the positive y direction. The projectile’s initial vertical velocity is $(27.0 m/s)(sin 45 ^{\circ})$. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. It starts at a height of y = 1.0 m, and the final vertical position is 13.0 m. Use equation 2.11b to find the time of flight of 3.108 s. The other root, a smaller one, is discarded. It corresponds to when the baseball reached y = 13.0 m on the way up, but the baseball landed on the roof on the way down. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(27.0 m/s)(cos 45 ^{\circ})(3.108 s) = 59.3 m.$$