/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 S In a local diner, a customer s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

S In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strikes the floor at distance \(d\) from the base of the counter. If the height of the counter is \(h\), (a) find an expression for the time \(t\) it takes the cup to fall to the floor in terms of the variables \(h\) and \(g\). (b) With what speed does the mug leave the counter? Answer in terms of the variables \(d, g\), and \(h .\) (c) In the same terms, what is the speed of the cup immediately before it hits the floor? (d) In terms of \(h\) and \(d\), what is the direction of the cup's velocity immediately before it hits the floor?

Short Answer

Expert verified
(a) The time it takes to reach the floor is given by \( t = \sqrt{ \frac{2h}{g} } \). (b) The mug leaves the counter with a speed of 0 m/s. (c) The speed of the mug just before hitting the floor is \( v = \sqrt{ 2gh } \). (d) The direction of the mug's velocity immediately before it hits the floor is \( \theta = tan^{-1}( \frac{d}{h}) \) degrees below the horizontal

Step by step solution

01

Time of Fall

Use the formula \( t = \sqrt{ \frac{2h}{g} } \) where \( t \) is the time, \( h \) is the height, and \( g \) is the acceleration due to gravity. This equation is derived from the formula for time in free fall.
02

Initial Speed

The initial speed of the cup as it leaves the counter is 0 as it only starts sliding, not launched with any initial speed.
03

Speed Before Impact

Use the formula \( v = \sqrt{ 2gh } \) where \( v \) is the speed, \( g \) is the gravitational acceleration and \( h \) is the height. This equation is derived from the formula for speed in free fall.
04

Calculate the Direction

To find the direction of the cup's velocity immediately before it hits the floor, use the formula \( \theta = tan^{-1}( \frac{d}{h}) \) where \( \theta \) is the angle with respect to the horizontal direction, \( d \) is the distance between the base of the counter and where the cup hits the ground, and \( h \) is the height of the counter.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Fall
When you slide a coffee cup off a counter, it's an example of free fall. Free fall is a special motion where an object moves under the influence of gravity alone, without any support or resistance from surfaces.

Free fall assumes the only force acting on the object is gravity, which causes the object to accelerate downwards.
  • In the absence of air resistance, the object doesn't experience any horizontal forces after it leaves the counter.
  • This means its vertical motion is solely influenced by gravity.
Consequently, if you want to determine how long it takes for the cup to hit the floor, you can use the formula: \( t = \sqrt{ \frac{2h}{g} } \). Here, \(t\) stands for time, \(h\) is the height from which it falls, and \(g\) is the gravitational acceleration.

The calculation assumes an ideal scenario where gravity is the only acting force. This lets us simplify complex motions into understandable equations.
Gravitational Acceleration
Gravity is an omnipresent force giving objects weight, pulling them towards the center of the Earth. On Earth's surface, the gravitational acceleration \(g\) is approximately \(9.81 \ m/s^2\).

The significance of gravitational acceleration in physics is profound because it allows us to predict and calculate free falling objects' motion.
  • It's the reason why all objects, when dropped from a height, accelerate at a uniform rate.
  • Gravitational acceleration is constant and does not depend on the object's mass, shape, or size.
In practical terms, when analyzing the motion of the coffee cup from the counter, \(g\) determines how quickly it speeds up on its descent. This gives us the speed just before impact through the equation:
\( v = \sqrt{2gh} \)
This formula factors in how height \(h\) and the conversion from potential to kinetic energy affect the cup's final speed.
Velocity Calculation
Velocity, in physics, isn’t just speed. It's speed with a direction. Calculating the velocity of our coffee cup is essential for understanding its behavior as it flies off the counter.
  • After the initial slide, gravity pulls the cup directly downwards while it continues to move horizontally.
  • The resulting velocity just before impact is a blend of its horizontal and vertical components.
Using the formula \( v = \sqrt{2gh} \), we found the vertical component, but we must also consider the horizontal distance \(d\).

To fully describe the cup's motion, you need to synthesize these components, resulting in two distinct speeds merging into one 'impact velocity.'

The horizontal velocity component remains constant, whereas the vertical velocity increases due to gravity, highlighting the complex interplay between different forces and directions.
Trigonometry in Physics
Determining the direction of the velocity vector as the cup strikes the floor involves a bit of trigonometry. This math branch helps us understand relationships between angles and distances in physics.

When the cup leaves the counter, its velocity vector consists of two components: horizontal and vertical. To find the angle \(\theta\) of the impact velocity with the horizontal, we use the tangent function:
\( \theta = \tan^{-1}(\frac{d}{h}) \).

This equation tells us the cup's trajectory as it falls:
  • The angle \(\theta\) depends on how far \(d\) it lands from the counter base and its fall height \(h\).
  • The further \(d\) is compared to \(h\), the smaller the angle, indicating a flatter descent.
Trigonometry aids in visualizing how the cup's motion changes direction when it collides with the ground, giving a complete picture of its movement path.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A man in a maze makes three consecutive displacements. His first displacement is \(8.00 \mathrm{~m}\) westward, and the second is \(13.0 \mathrm{~m}\) northward. At the end of his third displacement he is back to where he started. Use the graphical method to find the magnitude and direction of his third displacement.

A bomber is flying horizontally over level terrain at a speed of \(275 \mathrm{~m} / \mathrm{s}\) relative to the ground and at an altitude of \(3.00 \mathrm{~km}\). (a) The bombardier releases one bomb. How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call, "Bombs away!" Consequently, the pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where is the plane relative to the bomb's point of impact when the bomb hits the ground? (c) The plane has a telescopic bombsight set so that the bomb hits the target seen in the sight at the moment of release. At what angle from the vertical was the bombsight set?

A farm truck moves due east with a constant velocity of \(9.50 \mathrm{~m} / \mathrm{s}\) on a limitless, horizontal stretch of road. A boy riding on the back of the truck throws a can of soda upward (Fig. P3.54) and catches the projectile at the same location on the truck bed, but \(16.0\) m farther down the road. (a) In the frame of reference of the truck, at what angle to the vertical does the boy throw the can? (b) What is the initial speed of the can rela- tive to the truck? (c) What is the shape of the can's tra- jectory as seen by the boy? An observer on the ground watches the boy throw the can and catch it. In this observer's frame of reference, (d) describe the shape of the can's path and (e) determine the initial velocity of the can.

A hunter wishes to cross a river that is \(1.5 \mathrm{~km}\) wide and flows with a speed of \(5.0 \mathrm{~km} / \mathrm{h}\) parallel to its banks. The hunter uses a small powerboat that moves at a maximum speed of \(12 \mathrm{~km} / \mathrm{h}\) with respect to the water. What is the minimum time necessary for crossing?

A playground is on the flat roof of a city school, \(6.00 \mathrm{~m}\) above the street below (Fig. P3.34). The vertical wall of the building is \(h=7.00 \mathrm{~m}\) high, to form a l-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of \(\theta=53.0^{\circ}\) above the horizontal at a point \(d=24.0 \mathrm{~m}\) from the base of the building wall. The ball takes \(2.20\) s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.