/*! 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 45 Here's an interesting challenge ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Here's an interesting challenge you can give to a friend. Hold a $1 (or larger) bill by an upper corner. Have a friend prepare to pinch a lower corner, putting her fingers near but not touching the bill. Tell her to try to catch the bill when you drop it by simply closing her fingers. This seems like it should be easy, but it's not. After she sees that you have released the bill, it will take her about 0.25 s to react and close her fingers-which is not fast enough to catch the bill. How much time does it take for the bill to fall beyond her grasp? The length of a bill is 16 cm.

Short Answer

Expert verified
It takes approximately 0.181 seconds for the bill to fall beyond her grasp.

Step by step solution

01

Identify all known values

The length of the bill provides the distance \(d\) which is 16 cm. The acceleration due to gravity \(g\) is known and standard, typically rounded to approximately 9.8 m/s² (though in this problem, conversion is necessary to ensure the units match - the gravity value used here should be 980 cm/s² rather than m/s²).
02

Identify the unknown value

The problem asks for how much time it takes for the bill to fall beyond her grasp. Therefore, time (\(t\)) is the unknown value.
03

Setup and solve for unknown value

We set up the problem using the formula for distance travelled in free fall \(d = 0.5 * g * t^2\), and rearrange to solve for time, \(t = \sqrt{2d/g}\). Substituting known values into this equation, we get \(t = \sqrt{2 * 16cm / 980cm/s²}.\) Solving this gives us approximately \(t = 0.181 s.\)

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Key Concepts

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

Gravitational Acceleration
When we talk about objects falling freely towards the Earth, such as a dollar bill in our experiment, they are subject to a constant acceleration known as gravitational acceleration. This rate of acceleration is denoted by the symbol and has an approximate value of 9.8 m/s² on the surface of the Earth. Since acceleration due to gravity is constant, it plays a pivotal role in free fall problems.One crucial aspect to consider is the consistency of units when performing calculations. As seen in the original exercise, we converted distance to centimeters so it matches the unit of acceleration (980 cm/s²). Understanding that gravitational acceleration is the same for all objects regardless of their mass (neglecting air resistance) is key to solving free fall problems accurately. Simply put, if we dropped the bill alongside a heavier object from the same height, both would hit the ground at the same time.
Reaction Time
Reaction time is the interval between the moment when an event occurs and when the response to that event begins. In the case of our dollar bill exercise, the event is the release of the bill, and the response is the attempt to catch it. It is highlighted in the problem that the average human reaction time is about 0.25 seconds.Understanding reaction time is important because it can determine success or failure in time-sensitive activities, such as catching the bill. Knowing that most people cannot react rapidly enough to catch the falling bill demonstrates human limitations in processing speed and motor coordination. When teaching about reaction time, it is useful to compare it with different activities or scenarios, emphasizing its role in daily life as well as in physics experiments.
Kinematic Equations
The kinematic equations describe the motion of objects without considering the forces that cause this motion. They are a set of formulas that connect the five kinematic variables: displacement (\(d\)), initial velocity (\(v_i\)), final velocity (\(v_f\)), acceleration (\(a\)), and time (\(t\)). For free fall problems, the initial velocity is usually zero, making the kinematic equations simpler.In our exercise, we focused on the equation\, which relates distance fallen to gravitational acceleration and time. Now you're familiar with why the equation had to be rearranged to solve for time—displacement and acceleration were the known quantities. The ability to manipulate these equations is vital for students to solve a variety of motion-related problems, making them a fundamental aspect of physics education.

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Most popular questions from this chapter

II Haley is driving down a straight highway at 75 mph. A construction sign warns that the speed limit will drop to \(55 \mathrm{mph}\) in \(0.50 \mathrm{mi} .\) What constant acceleration (in \(\mathrm{m} / \mathrm{s}\) ) will bring Haley to this lower speed in the distance available?

A truck driver has a shipment of apples to deliver to a destination 440 miles away. The trip usually takes him 8 hours. Today he finds himself daydreaming and realizes 120 miles into his trip that he is running 15 minutes later than his usual pace at this point. At what speed must he drive for the remainder of the trip to complete the trip in the usual amount of time?

When jumping, a flea reaches a takeoff speed of \(1.0 \mathrm{m} / \mathrm{s}\) over a distance of \(0.50 \mathrm{mm}\) a. What is the flea's acceleration during the jump phase? b. How long does the acceleration phase last? c. If the flea jumps straight up, how high will it go? (Ignore air resistance for this problem; in reality, air resistance plays a large role, and the flea will not reach this height.)

Certain insects can achieve seemingly impossible accelerations while jumping. The click beetle accelerates at an astonishing \(400 \mathrm{g}\) over a distance of \(0.60 \mathrm{cm}\) as it rapidly bends its thorax, making the "click" that gives it its name. a. Assuming the beetle jumps straight up, at what speed does it leave the ground? b. How much time is required for the beetle to reach this speed? c. Ignoring air resistance, how high would it go?

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket's acceleration?

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