Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is h. The mug slides off the counter and strikes the floor at distance dfrom the base of the counter.

(a) With what velocity did the mug leave the counter?

(b) What was the direction of the mug's velocity just before it hit the floor?

Short answer

1. The velocity of the mug when it leaves the counter is $d·\sqrt{\frac{g}{2·h}}$.
2. The direction of the mug’s velocity just before it hit the floor is $\frac{2·h}{d}$.

Step-by-step solution

Given Data

The height of the counter is h and the mug slides off the counter and strikes the floor at distance d from the base of the counter.

Definition of velocity

The rate of change of an object's location with regard to a frame of reference is its velocity, which is a function of time. A statement of an object's speed and direction of motion is referred to as velocity.

Find the velocity of the mug when it leaves the counter.

(a)

The initial angle is ${\theta }_0=0$and the initial height is $y_0=h$.

Therefore,

Find the direction of the mug’s velocity just before it hit the floor. 

(b)

The $v_x$component of motion is constant in horizontal projectile motion, but the $v_y$component rises in magnitude:

$$\begin{gathered} v_x=v_0 \\ {v}_y=\sqrt{2·g·y} \end{gathered}$$

The mug's velocity components at the instant of collision with the ground are:

$$\begin{gathered} v_x=v_0 \\ =d·\sqrt{\frac{g}{2·h}} \\ {v}_y=\sqrt{2·g·h} \end{gathered}$$

The velocity angle $\theta$at the instant of contact may be computed as follows:

$$\begin{gathered} \tan \theta =\frac{v_y}{v_x} \\ =\frac{\sqrt{2·g·h}}{d·\sqrt{\frac{g}{2·h}}} \\ =\frac{2·h}{d} \end{gathered}$$