/*! 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 92 Two students are asked to find t... [FREE SOLUTION] | 91Ó°ÊÓ

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Two students are asked to find the height of a particular building using a barometer. Instead of using the barometer as an altitude-measuring device, they take it to the roof of the building and drop it off, timing its fall. One student reports a fall time of \(2.0 \mathrm{~s}\), and the other, \(2.3 \mathrm{~s}\). What \(\%\) difference does the 0.3 s make for the estimates of the building's height?

Short Answer

Expert verified
The 0.3 s difference results in a 32.3% change in height estimation.

Step by step solution

01

Understanding the Problem

Our task is to evaluate the height of the building using direct measurements of fall time to calculate the impact of a 0.3 s difference on height estimation.
02

Using Free-Fall Formula

We'll calculate the height for each time interval using the formula for the distance fallen under gravity: \[ h = \frac{1}{2} g t^2 \]where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( t \) is the fall time.
03

Height Calculation for 2.0 Seconds

Substitute \( t = 2.0 \, \text{s} \) into the formula: \[ h = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (2.0)^2 = 19.6 \, \text{m} \]
04

Height Calculation for 2.3 Seconds

Substitute \( t = 2.3 \, \text{s} \) into the formula: \[ h = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (2.3)^2 = 25.921 \, \text{m} \]
05

Calculate Difference and Percentage Difference

The difference in height is \[ 25.921 \, \text{m} - 19.6 \, \text{m} = 6.321 \, \text{m} \]The percentage difference relative to the first measurement is \[ \frac{6.321}{19.6} \times 100\% = 32.3\% \]

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Key Concepts

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

Acceleration Due to Gravity
Gravity is a fundamental force that attracts objects with mass toward each other. On Earth, this force gives objects a downward acceleration. The value of this acceleration, denoted as \( g \), is approximately \( 9.8 \, \text{m/s}^2 \). This means that if you drop an object, it will accelerate downwards at a rate of \( 9.8 \, \text{m/s}^2 \) for every second it is falling.
Understanding this concept is crucial when dealing with physics problems that involve objects in free fall. The constant rate of acceleration due to gravity allows us to make predictions about how fast an object will be moving at any given time and the distance it will travel during its fall.
  • If no other forces are acting on an object, it is said to be in free fall.
  • The acceleration due to gravity is constant regardless of the mass of the falling object.
  • This acceleration affects the velocity and distance of the object over time.
Using this concept, we can predict movements and make calculations such as estimating the height from which an object is falling using the time it takes to reach the ground.
Height Estimation
Estimating the height of an object, such as a building, using a dropped object is a practical application of physics. The key formula used here is derived from the equations of motion, specifically for free fall: \[ h = \frac{1}{2} g t^2 \]This formula lets us calculate the distance \( h \), which is the height the object fell, based on the time \( t \) it takes to fall. Here’s how it works:
  • Consider a known acceleration \( g = 9.8 \, \text{m/s}^2 \).
  • Measure how long it takes for the object to hit the ground (time \( t \)).
  • Substitute these values into the formula to find the height \( h \).
For example, if a barometer is dropped and it takes \( 2.0 \, \text{s} \) to fall, substituting \( t = 2.0 \) into the formula gives us a height of \( 19.6 \, \text{m} \). If the time is \( 2.3 \, \text{s} \), the height would be \( 25.921 \, \text{m} \), showing how time influences the height that is estimated.
Percentage Difference Calculation
Understanding and calculating the percentage difference is important when comparing two measurements or estimates. It is a measure of how much one value deviates from another in terms of percentage. This can be particularly useful in experiments and real-world applications to assess the accuracy of a measurement. Here is how you calculate it:
  • Determine the difference between two values. For instance, two height measurements (\( 25.921 \, \text{m} \) and \( 19.6 \, \text{m} \)) have a difference of \( 6.321 \, \text{m} \).
  • Take the second value and divide the difference by the original value (\( 19.6 \, \text{m} \)).
  • Multiply the result by 100 to get the percentage difference.
The formula is represented as:\[ \text{Percentage Difference} = \left( \frac{\text{Difference}}{\text{Original Value}} \right) \times 100\% \]In our case, this comes out to be \( 32.3\% \), meaning the second measurement is 32.3% greater than the original measurement. Such calculations are very helpful in identifying discrepancies or variations in data, thus, providing insight into the reliability and accuracy of studies or experiments.

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Most popular questions from this chapter

(III) A rock is thrown vertically upward with a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) Exactly \(1.00 \mathrm{~s}\) later, a ball is thrown up vertically along the same path with a speed of \(18.0 \mathrm{~m} / \mathrm{s}\). \((a)\) At what time will they strike each other? \((b)\) At what height will the collision occur? (c) Answer \((a)\) and \((b)\) assuming that the order is reversed: the ball is thrown \(1.00 \mathrm{~s}\) before the rock.

(I) A sprinter accelerates from rest to \(9.00 \mathrm{~m} / \mathrm{s}\) in \(1.28 \mathrm{~s}\). What is her acceleration in (a) \(\mathrm{m} / \mathrm{s}^{2} ;\) (b) \(\mathrm{km} / \mathrm{h}^{2}\) ?

Agent Bond is standing on a bridge, \(13 \mathrm{~m}\) above the road below, and his pursuers are getting too close for comfort. He spots a flatbed truck approaching at \(25 \mathrm{~m} / \mathrm{s}\), which he measures by knowing that the telephone poles the truck is passing are \(25 \mathrm{~m}\) apart in this country. The bed of the truck is \(1.5 \mathrm{~m}\) above the road, and Bond quickly calculates how many poles away the truck should be when he jumps down from the bridge onto the truck, making his getaway. How many poles is it?

(II) A stone is thrown vertically upward with a speed of \(24.0 \mathrm{~m} / \mathrm{s}\). \((a)\) How fast is it moving when it reaches a height of \(13.0 \mathrm{~m} ?(b)\) How much time is required to reach this height? (c) Why are there two answers to (b)?

(II) According to a rule-of-thumb, every five seconds between a lightning flash and the following thunder gives the distance to the flash in miles. Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in \(\mathrm{m} / \mathrm{s}\) from this rule. What would be the rule for kilometers?

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