Question

a bartender slides a beer mug at 1.6 m/s towards a customer at the end of a frictionless bar that is 1.5 m tall. the customer makes a grab for the mug and misses, and the mug sails off the end of the bar. (a) how far away from the end of the bar does the mug hit the floor? (b) what are the speed and direction of the mug at impact? speed m/s direction ° below the horizontal

Explanation:

Step1: Find time of fall

The mug is in free - fall in the vertical direction. The height of the bar is $h = 1.5\ m$, and the initial vertical velocity $v_{0y}=0\ m/s$. Using the equation $h=v_{0y}t+\frac{1}{2}gt^{2}$, since $v_{0y} = 0$, we have $h=\frac{1}{2}gt^{2}$. Solving for $t$, we get $t=\sqrt{\frac{2h}{g}}$. Substituting $h = 1.5\ m$ and $g = 9.8\ m/s^{2}$, we have $t=\sqrt{\frac{2\times1.5}{9.8}}\ s$.
$$t=\sqrt{\frac{3}{9.8}}\ s\approx0.55\ s$$

Step2: Find horizontal distance

The horizontal velocity $v_{x}=1.6\ m/s$ (constant because there is no horizontal acceleration in the absence of friction). Using the equation $x = v_{x}t$, substituting $v_{x}=1.6\ m/s$ and $t = 0.55\ s$, we get $x=1.6\times0.55\ m = 0.88\ m$.

Step3: Find vertical velocity at impact

Using the equation $v_{y}=v_{0y}+gt$, with $v_{0y}=0\ m/s$, $g = 9.8\ m/s^{2}$ and $t = 0.55\ s$, we have $v_{y}=0 + 9.8\times0.55\ m/s=5.39\ m/s$.

Step4: Find speed at impact

The speed $v$ at impact is given by the Pythagorean theorem $v=\sqrt{v_{x}^{2}+v_{y}^{2}}$. Substituting $v_{x}=1.6\ m/s$ and $v_{y}=5.39\ m/s$, we get $v=\sqrt{1.6^{2}+5.39^{2}}\ m/s=\sqrt{2.56 + 29.0521}\ m/s=\sqrt{31.6121}\ m/s\approx5.62\ m/s$.

Step5: Find direction at impact

The direction $\theta$ below the horizontal is given by $\tan\theta=\frac{v_{y}}{v_{x}}$. Substituting $v_{x}=1.6\ m/s$ and $v_{y}=5.39\ m/s$, we have $\theta=\arctan(\frac{5.39}{1.6})\approx73.5^{\circ}$.

Answer:

(a) $0.88$
(b) speed: $5.62$
direction: $73.5$