Question

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug slides off the counter and strikes the floor 2.00 m from the base of the counter. (a) With what velocity did the mug leave the counter? m/s (b) What was the direction of the mug's velocity just before it hit the floor? °

Answer 1

Answer:

a) $V_{x}=3.72m/s$, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

$y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}$

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

$0=y_{0}+\frac{1}{2}a_{y}t^{2}$

so we can solve for t, so we get:

$t=\sqrt{\frac{-(2)y_{0}}{a}}$

so now we can substitute the known values, so we get:

$t=\sqrt{\frac{-(2)(1.42)}{-9.8}}$

which yields:

t=0.538s

So we can use this value to find the velocity in x:

$V_{x}=\frac{x}{t}$

When substituting we get:

$V_{x}=\frac{2m}{0.538s}$

which yields:

$V_{x}=3.72m/s$

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

$\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

We know that $V_{0}$ is zero, so we can simplify the expression:

$\Delta y=\frac{V_{yf}^{2}}{2a}$

So we can solve the equation for $V_{yf}^{2}$ so we get:

$V_{yf}=\sqrt{2\Delta y a}$

and when substituting the known values we get:

$V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}$

which yields:

$V_{yf}=-5.28m/s$

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

$tan \theta =\frac{V_{y}}{V_{x}}$

when solving for theta we get:

$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$

We can substitute so we get:

$\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})$

which yields:

$\theta = -54.83^{o}$