3. In a local diner, a customer slides an empty coffee mug down the counter for a refill. The mug slides off the edge of the counter and crashes into the floor at a distance d from the base of the counter. The counter's height is h. The following answers should be only in terms of d, h, and g. a) What is the y-component of the mug's velocity vector just after it slides off the counter? b) Using the answer to (a), determine the amount of time the mug is airborne. c) What is the x-component of the mug's velocity vector just as it slides off the counter's edge? Hint: this is just like Part A of Lab #2.
3. In a local diner, a customer slides an empty coffee mug down the counter for a refill. The mug slides off the edge of the counter and crashes into the floor at a distance d from the base of the counter. The counter's height is h. The following answers should be only in terms of d, h, and g. a) What is the y-component of the mug's velocity vector just after it slides off the counter? b) Using the answer to (a), determine the amount of time the mug is airborne. c) What is the x-component of the mug's velocity vector just as it slides off the counter's edge? Hint: this is just like Part A of Lab #2.
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![**Physics Problem: Projectile Motion of a Coffee Mug**
In a local diner, a customer slides an empty coffee mug down the counter for a refill. The mug slides off the edge of the counter and crashes into the floor at a distance \( d \) from the base of the counter. The counter's height is \( h \). The following answers should be only in terms of \( d \), \( h \), and \( g \).
a) **Find the Y-Component of the Mug’s Velocity Vector Just After It Slides Off the Counter:**
The initial vertical velocity component \( v_{y0} \) of the mug is 0 since it starts with horizontal motion only.
b) **Determine the Amount of Time the Mug is Airborne:**
The time \( t \) the mug is airborne can be found using the following equation for free fall:
\[ h = \frac{1}{2} g t^2 \]
Solve for \( t \):
\[ t = \sqrt{\frac{2h}{g}} \]
c) **Find the X-Component of the Mug’s Velocity Vector Just as It Slides Off the Counter’s Edge:**
Hint: Use principles from Part A of Lab #2. The horizontal velocity \( v_{x0} \) can be found from:
\[ d = v_{x0} \cdot t \]
\[ v_{x0} = \frac{d}{t} \]
d) **What is the Mug’s Impact Velocity Vector?**
The impact velocity \( v \) is a combination of both horizontal and vertical components:
\[ v_{x} = v_{x0} \]
\[ v_{y} = g \cdot t \]
The magnitude of the velocity vector is:
\[ v = \sqrt{v_{x}^2 + v_{y}^2} \]
e) **Determine the Angle at Which the Mug Hits the Floor:**
Use the inverse tangent function to find the angle \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) \]
This analysis breaks down the projectile motion of the mug into basic components using physics principles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c5a5d39-4610-4542-84f3-e3a5fb07be7c%2Ff3eef280-17ad-416f-84f1-693e12866e9d%2Fs0bfnqw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Projectile Motion of a Coffee Mug**
In a local diner, a customer slides an empty coffee mug down the counter for a refill. The mug slides off the edge of the counter and crashes into the floor at a distance \( d \) from the base of the counter. The counter's height is \( h \). The following answers should be only in terms of \( d \), \( h \), and \( g \).
a) **Find the Y-Component of the Mug’s Velocity Vector Just After It Slides Off the Counter:**
The initial vertical velocity component \( v_{y0} \) of the mug is 0 since it starts with horizontal motion only.
b) **Determine the Amount of Time the Mug is Airborne:**
The time \( t \) the mug is airborne can be found using the following equation for free fall:
\[ h = \frac{1}{2} g t^2 \]
Solve for \( t \):
\[ t = \sqrt{\frac{2h}{g}} \]
c) **Find the X-Component of the Mug’s Velocity Vector Just as It Slides Off the Counter’s Edge:**
Hint: Use principles from Part A of Lab #2. The horizontal velocity \( v_{x0} \) can be found from:
\[ d = v_{x0} \cdot t \]
\[ v_{x0} = \frac{d}{t} \]
d) **What is the Mug’s Impact Velocity Vector?**
The impact velocity \( v \) is a combination of both horizontal and vertical components:
\[ v_{x} = v_{x0} \]
\[ v_{y} = g \cdot t \]
The magnitude of the velocity vector is:
\[ v = \sqrt{v_{x}^2 + v_{y}^2} \]
e) **Determine the Angle at Which the Mug Hits the Floor:**
Use the inverse tangent function to find the angle \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) \]
This analysis breaks down the projectile motion of the mug into basic components using physics principles.
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Transcribed Image Text:**d) What is the mug’s impact velocity vector?**
**e) Determine the angle at which the mug hits the floor. Hint: this is a simple vector idea applied to (d).**
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