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13P

Expert-verifiedFound in: Page 150

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.
- 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,

$x={v}_{0}\xb7t\phantom{\rule{0ex}{0ex}}y=h-\frac{1}{2}\xb7g\xb7{t}^{2}\phantom{\rule{0ex}{0ex}}$

The maximum distance is ${x}_{\mathrm{max}}=d$ and the final height is $y=0$:

$d={v}_{0}\xb7{t}_{\mathrm{max}}\phantom{\rule{0ex}{0ex}}0=h-\frac{1}{2}\xb7g\xb7{t}_{{\mathrm{max}}^{2}}\phantom{\rule{0ex}{0ex}}$

Rewrite the equations above:

${v}_{0}=\frac{d}{{t}_{\mathrm{max}}}\phantom{\rule{0ex}{0ex}}{t}_{\mathrm{max}}=\sqrt{\frac{2\xb7h}{g}}\phantom{\rule{0ex}{0ex}}$

Substitute the second equation into first and calculate the result:

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

(b)

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

width="94" style="max-width: none; vertical-align: -41px;" ${v}_{x}={v}_{0}\phantom{\rule{0ex}{0ex}}{v}_{y}=\sqrt{2\xb7g\xb7y}\phantom{\rule{0ex}{0ex}}$

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

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

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

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