Suggested languages for you:

Americas

Europe

14P

Expert-verifiedFound in: Page 78

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**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 d ****from 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?**

- The velocity of the mug when it leaves the counter is $d\xb7\sqrt{\frac{g}{2\xb7h}}$.
- The direction of the mug’s velocity just before it hit the floor is $\frac{2\xb7h}{d}$.

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**.**

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.

(a)

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

Therefore,

(b)

The ${v}_{x}$component of motion is constant in horizontal projectile motion, but the ${v}_{y}$component rises in magnitude:

${v}_{x}={v}_{0}\phantom{\rule{0ex}{0ex}}{v}_{y}=\sqrt{2\xb7g\xb7y}\phantom{\rule{0ex}{0ex}}$

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

${v}_{x}={v}_{0}\phantom{\rule{0ex}{0ex}}=d\xb7\sqrt{\frac{g}{2\xb7h}}\phantom{\rule{0ex}{0ex}}{v}_{y}=\sqrt{2\xb7g\xb7h}$

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

$\mathrm{tan}\theta =\frac{{v}_{y}}{{v}_{x}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2\xb7g\xb7h}}{d\xb7\sqrt{\frac{g}{2\xb7h}}}\phantom{\rule{0ex}{0ex}}=\frac{2\xb7h}{d}\phantom{\rule{0ex}{0ex}}$

94% of StudySmarter users get better grades.

Sign up for free