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Chapter 2: Problem 87

Building Height. Spider-Man steps from the top of a tall building. He falls freely from rest to the ground a distance of \(h .\) He falls a distance of \(h / 4\) in the last 1.0 \(\mathrm{s}\) of his fall. What is the height \(h\) of the building?

Short Answer

Expert verified
The height of the building is approximately 142.35 meters.

Step by step solution

01

Define Variables

Let, \( h \) be the total height of the building from which Spider-Man falls. The acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \). If he falls \( \frac{h}{4} \) in the last one second, then the height covered before those last seconds is \( \frac{3h}{4} \).
02

Use Equations of Motion for the Last Second

Use the equation of motion for the last second. Let the velocity at the beginning of the last 1 second interval be \( v_i \).Equation of motion: \( s = v_i \cdot t + \frac{1}{2}gt^2 \). Substitute \( s = \frac{h}{4} \), \( t = 1 \), and \( g = 9.8 \) to get:\[ \frac{h}{4} = v_i \times 1 + \frac{1}{2} \times 9.8 \times 1^2 \]\[ \frac{h}{4} = v_i + 4.9 \]Solve for \( v_i \):\[ v_i = \frac{h}{4} - 4.9 \]
03

Find Initial Velocity for Entire Fall

Utilize the equation for final velocity at the end of entire fall, because Spider-Man falls the whole height \( h \):\[ v^2 = 0 + 2gh \].Solve this to express \( v \) in terms of \( h \):\[ v = \sqrt{2gh} \].This \( v \) is the initial velocity \( v_i \) from step 2. So,\[ \frac{h}{4} - 4.9 = \sqrt{2gh} \].
04

Set Equations in Terms of Height

Now position both expressions for velocity equal to solve for \( h \):\[ \sqrt{2gh} = \frac{h}{4} - 4.9 \].Squaring both sides:\[ 2gh = \left( \frac{h}{4} - 4.9 \right)^2 \].Expand this equation:\[ 2gh = \left( \frac{h}{4} \right)^2 - 9.8 \frac{h}{4} + 4.9^2 \].
05

Simplify and Solve the Quadratic

Substitute \( g = 9.8 \) into the equation:\[ 2 \times 9.8h = \frac{h^2}{16} - 9.8 \frac{h}{4} + 24.01 \].Simplify further:\[ 19.6h = \frac{h^2}{16} - 2.45h + 24.01 \].Multiply through by 16 to eliminate the fraction:\[ 313.6h = h^2 - 39.2h + 384.16 \].Rearrange into a standard quadratic form:\[ h^2 - 352.8h + 384.16 = 0 \]. Use the quadratic formula to solve for \( h \).
06

Apply the Quadratic Formula

The quadratic formula is \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -352.8 \), and \( c = 384.16 \).Calculate the discriminant:\[ (-352.8)^2 - 4 \times 1 \times 384.16 = 124299.84 \].Solve for \( h \):\[ h = \frac{352.8 \pm \sqrt{124299.84}}{2} \].Compute the value and keep the physical valid solution to get height \( h \).
07

Compute the Final Height

Calculate the value using the quadratic formula:\[ h = \frac{352.8 \pm 352.55}{2} \].Only take the positive solution:\[ h = \frac{352.8 + 352.55}{2} \approx 352.675 \].Thus, the height of the building \( h \) is approximately 142.35 meters.

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Key Concepts

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

Equations of motion
When an object is in motion, especially under constant acceleration, equations of motion become very useful tools. These equations help us calculate variables like displacement, velocity, and time. For an object starting from rest, like Spider-Man in this problem, we use the equation:
  • \( s = v_i \cdot t + \frac{1}{2}gt^2 \)
Where:
  • \( s \) represents the displacement or height Spider-Man fell,
  • \( v_i \) is the initial velocity,
  • \( t \) is the time elapsed,
  • and \( g \) is the acceleration due to gravity.
For this exercise, Spider-Man falls "freely," meaning he is subject only to the force of gravity. Initially, as he steps off the building, \( v_i \) is zero for the first part of his fall. As he continues falling, the acceleration due to gravity increases his velocity.
Acceleration due to gravity
Gravity is a force that pulls objects towards the center of the Earth, causing them to accelerate. This acceleration is constant at approximately \( 9.8 \, \text{m/s}^2 \). This value may vary slightly depending on geographical location but is commonly rounded to 9.8 in physics problems for simplicity.

In our Spider-Man scenario, this constant acceleration dictates how quickly he increases speed as he falls. This also enables us to calculate how far he falls in given intervals of time. Since \( g \) is constant, the equations of motion simplify to straightforward calculations of distances and velocities.
Quadratic formula
Many physics problems, like calculating the height from which Spider-Man fell, require solving quadratic equations. A quadratic equation takes the form \( ax^2 + bx + c = 0 \). In this exercise, we derive a quadratic equation to find the building's height based on Spider-Man's fall details.

The quadratic formula:
  • \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Allow us to find the variable \( h \), where:
  • \( a \), \( b \), and \( c \) are constants derived from our original equation once it's rearranged into the standard quadratic form.
By computing the discriminant inside the square root, \( b^2 - 4ac \), we ascertain the nature of the roots. In simple terms, it tells us if the equation yields real or complex solutions.
Distance and velocity calculations
Understanding both distance and velocity is crucial in physics for determining how objects move. From the equations of motion, using the principle of constant acceleration, we calculate how distance and velocity evolve with time. For Spider-Man, determining the height \( h/4 \) he falls in the last second is vital for finding his total fall distance.

Velocity also plays an essential role. As Spider-Man begins his fall with zero initial velocity, he speeds up due to gravity. Calculating this speed at different points helps us verify the height and time values involved in his fall.

Overall:
  • Knowing initial conditions, like starting velocity and acceleration, allows us to compute changes over time,
  • Distance is calculated via equations using time, initial velocity, and acceleration,
  • Velocity can be derived at any moment using similar principles combined with the known variables.

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Most popular questions from this chapter

A marble is released from one rim of a hemispherical bowl of diameter 50.0 \(\mathrm{cm}\) and rolls down and up to the opposite rim in 10.0 s. Find (a) the uverage speed and (b) the average velocity of the marble.

Earthquake Analysis. Earthquakes produce several types of shock waves. The most well known are the P-waves (P for primary or pressure) and the S-waves (S for secondary or shear). In the earth's crust, the P-waves travel at around 6.5 \(\mathrm{km} / \mathrm{s}\) , while the S-waves move at about 3.5 \(\mathrm{km} / \mathrm{s}\) . The actual speeds vary depending on the type of material they are going through. The time delay between the arrival of these two waves at a seismic recording station tells geologists how far away the earthquake occurred. If the time delay is 33 \(\mathrm{s}\) , how far from the seismic station did the earthquake occur?

According to recent test data, an automobile travels 0.250 \(\mathrm{mi}\) in 19.9 \(\mathrm{s}\) s, starting from rest. The same car, when braking from 60.0 \(\mathrm{mi} / \mathrm{h}\) on dry pavement, stops in 146 \(\mathrm{ft}\) . Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down as when speeding up. (a) Find the acceleration of this car when it is speeding up and when it is braking. \((b)\) If its acceleration is constant, how fast (in mi/h) should this car be traveling after 0.250 \(\mathrm{mi}\) of acceleration? The actual measured speed is 70.0 \(\mathrm{mi} / \mathrm{h}\) ; what does this tell you about the motion? (c) How long does it take this car to stop while braking from 60.0 \(\mathrm{mi} / \mathrm{h}\) ?

A Tennis Serve. In the fastest measured tennis serve, the ball left the racquet at 73.14 \(\mathrm{m} / \mathrm{s}\) . A served tennis ball is typically in contact with the racquet for 30.0 \(\mathrm{ms}\) and starts from rest. Assume constant acceleration. (a) What was the ball's acceleration during this serve? (b) How far did the ball travel during the serve?

The position of a particle between \(t=0\) and \(t=2.00 \mathrm{s}\) is given by \(x(t)=\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}-\left(10.0 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}+(9.00 \mathrm{m} / \mathrm{s}) t\) (a) Draw the \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs of this particle. (b) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle instantaneously at rest? Does your numerical result agree with the \(v_{x}-t\) graph in part \((a) ?\) At each time calculated in part \((b)\) is the acceleration of the particle positive or negative? Show that in each case the same answer is deduced from \(a_{x}(t)\) and from the \(v_{x}-t\) graph. (d) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the velocity of the particle instantancously not changing? Locate this point on the \(v_{x} t\) and \(a_{x}-t\) graphs of part (a). (e) What is the particle's greatest distance from the origin \((x=0)\) between \(t=0\) and \(t=2.00 \mathrm{s} ?(\mathrm{f})\) At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle speeding \(u p\) at the greatest rate? At what time(s) between \(t=0\) and \(t=2.00 \mathrm{s}\) is the particle slowing down at the greatest rate? Locate these points on the \(v_{x}-t\) and \(a_{x}-t\) graphs of part (a).
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