Question

1. in a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.34 m. the mug slides off the counter and strikes the floor 1.20 m from the base of the counter (a) with what velocity did the mug leave the counter?

Explanation:

Step1: Analyze vertical - motion

The mug is in free - fall in the vertical direction. The initial vertical velocity $v_{0y}=0\ m/s$, the acceleration due to gravity $g = 9.8\ m/s^{2}$, and the vertical displacement $y=- 1.34\ m$ (taking downwards as negative). Use the equation $y=v_{0y}t+\frac{1}{2}at^{2}$. Since $v_{0y} = 0$, we have $y=\frac{1}{2}at^{2}$, so $t=\sqrt{\frac{-2y}{g}}$.
$$t=\sqrt{\frac{-2\times(-1.34)}{9.8}}=\sqrt{\frac{2.68}{9.8}}\approx0.52\ s$$

Step2: Analyze horizontal - motion

In the horizontal direction, there is no acceleration ($a_x = 0$), and the horizontal displacement $x = 1.20\ m$. The horizontal velocity $v_x$ is constant, and we use the equation $x = v_x t$. Then $v_x=\frac{x}{t}$.
Substitute $x = 1.20\ m$ and $t\approx0.52\ s$ into the formula: $v_x=\frac{1.20}{0.52}\approx2.29\ m/s$.

Answer:

$2.29\ m/s$