a baseball is launched horizontally from a height of 1.8 m. the baseball travels 0.50 m before hitting the…

Explanation:

Step1: Calculate time of fall

The vertical - motion of the baseball is a free - fall motion. The initial vertical velocity (v_{0y}=0\ m/s), the acceleration due to gravity (g = 9.8\ m/s^{2}), and the vertical displacement (y=- 1.8\ m) (taking downwards as negative). Using the equation (y = v_{0y}t+\frac{1}{2}at^{2}), since (v_{0y} = 0\ m/s), we have (y=\frac{1}{2}at^{2}), where (a=-g). So, (t=\sqrt{\frac{-2y}{g}}). [t=\sqrt{\frac{-2\times(-1.8)}{9.8}}=\sqrt{\frac{3.6}{9.8}}\approx0.606\ s]

Step2: Calculate horizontal velocity

The horizontal displacement (x = 0.50\ m). In horizontal motion (no acceleration, (a_x = 0)), the horizontal velocity (v_x) is constant and given by (v_x=\frac{x}{t}). [v_x=\frac{0.50}{0.606}\approx0.82\ m/s]

Answer:

(0.82)