Question

a baseball is launched horizontally from a height of 1.8 m. the baseball travels 0.50 m before hitting the ground. how fast is the baseball moving horizontally, rounded to the nearest hundredth? m/s

Explanation:

Step1: Analyze vertical - motion

The vertical - motion of the baseball is a free - fall motion. The initial vertical velocity \(v_{0y}=0\ m/s\), the acceleration \(a = g=9.8\ m/s^{2}\), and the vertical displacement \(y=- 1.8\ m\) (taking downwards as negative). Use the equation \(y = v_{0y}t+\frac{1}{2}at^{2}\). Since \(v_{0y} = 0\ m/s\), the equation simplifies to \(y=\frac{1}{2}at^{2}\).

$$t=\sqrt{\frac{- 2y}{a}}=\sqrt{\frac{-2\times(-1.8)}{9.8}}$$

Step2: Calculate the time of flight

$$t=\sqrt{\frac{3.6}{9.8}}\approx\sqrt{0.3673}\approx0.606\ s$$

Step3: Analyze horizontal - motion

In horizontal motion (where there is no acceleration, \(a_x = 0\)), the horizontal displacement \(x = 0.50\ m\). The horizontal velocity \(v_x\) is constant, and we use the equation \(x = v_x t\). Then \(v_x=\frac{x}{t}\).

$$v_x=\frac{0.50}{0.606}\approx0.82\ m/s$$

Answer:

\(0.82\)