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Chapter 2: Problem 61

A steel ball is dropped from a building's roof and passes a window, taking \(0.125 \mathrm{~s}\) to fall from the top to the bottom of the window, a distance of \(1.20 \mathrm{~m} .\) It then falls to a sidewalk and bounces back past the window, moving from bottom to top in \(0.125\) s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is \(2.00 \mathrm{~s}\). How tall is the building?

Short Answer

Expert verified
The building is approximately 25.87 meters tall.

Step by step solution

01

Analyze Information Given

We are given that the steel ball takes \(0.125\) s to fall \(1.20\) m from the top to the bottom of the window. It also takes \(0.125\) s to return to the top of the window upon bouncing back.The ball spends \(2.00\) s below the window from when it leaves to when it returns.
02

Calculate the Falling Speed Through the Window

We know the distance and the time it takes to pass the window. The distance is \(1.20\) m, and the time is \(0.125\) s. To find the speed at which the ball passes the window top, we use the formula:\[ v = \frac{distance}{time} = \frac{1.20 \, \text{m}}{0.125 \, \text{s}} = 9.6 \, \text{m/s} \] However, since it is both the entrance and the exit speed through the window due to symmetry, solve for the speed of the ball as it starts to pass the window.
03

Calculate Initial Speed Before Window

The speed at the top of the window, struck by the ball when returning, is the same as when it crosses going down. Using the equation for uniform accelerated motion:\[ v_f^2 = v_i^2 + 2ad \]where \(v_f = 9.6 \, \text{m/s}\), \(a = 9.8 \,\text{m/s}^2\), and \(d = 1.20 \, \text{m}\), solve for \(v_i\).\[ (9.6)^2 = v_i^2 + 2 \times 9.8 \times 1.20 \] \[ v_i^2 = 92.16 - 23.52 \] \[ v_i = \sqrt{68.64} \approx 8.28 \, \text{m/s} \]
04

Find Time Taken During Falling and Rising

Since it takes \(0.125\) s to fall through and the same to rise through the window, the ball spends 2.00 seconds below it total. First, calculate the time to fall from the roof to the top of the window:The total time from the roof to the sidewalk and back to the window is: \[ t_{total} = t_{falling\_window} + t_{below\_window} = 0.125 + 2.00 + 0.125 = 2.25 \, \text{s} \]
05

Calculate the Height of the Building

Now, the roof's height can be calculated using:\[ h = h_{window} + \frac{1}{2}gt_{roof}^2 \] First calculate \(t_{roof}\):\[ \frac{1}{2} gt_{roof}^2 = \frac{1}{2} \times 9.8 \times 2.25^2 = 24.67\] mAnd since\[ h_{window} = 1.20 \, \text{m} \] \[ h = x + 24.67 = 25.87 \, \text{m}\]
06

Conclude Total Building Height

Hence the total height of the building from which the ball was dropped:\[ h = 25.87 \, \text{m} \]

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Key Concepts

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

Free Fall
In physics, free fall describes the motion of an object under the influence of gravitational force only. An object in free fall experiences no other forces, such as friction or air resistance.

When a steel ball is dropped from the rooftop of a building, it enters a state of free fall, accelerating downward solely due to gravity. This acceleration occurs until the object is stopped by another force, like the sidewalk or, in our case, bounces back up.
  • During free fall, all objects accelerate at the same rate, regardless of their mass, because gravity is acting uniformly on them.
  • This rate of acceleration is constant and known to be approximately \(9.8\, \text{m/s}^2\) on Earth.
Understanding free fall is essential in calculating the building's total height in the given exercise because it relates to how long it takes the steel ball to fall past the window and then the entire building's height.
Acceleration due to Gravity
Acceleration due to gravity is a fundamental concept when dealing with objects in motion, particularly those in free fall. It defines how quickly an object speeds up as it falls under gravitational pull.

On Earth, this acceleration is approximately \(9.8\, \text{m/s}^2\). This constant rate means that with every passing second, the velocity of a freely falling object increases by \(9.8\, \text{m/s}\).
  • This measurement is crucial when calculating how fast the steel ball moves as it passes by the window.
  • It is also used to compute the time an object takes to fall from a certain height and its impact velocity.
In the exercise, acceleration due to gravity allows us to determine the ball's velocities and calculate the height of the building effectively.
Projectile Motion
Projectile motion combines vertical and horizontal movements of an object. In the exercise context, it applies to the ball's movement when it bounces and travels back past the window.

When analyzing projectile motion, it's essential to separate vertical and horizontal components, especially since gravity influences only the vertical motion.
  • The upward and downward motions of the ball have symmetrical properties, meaning the time taken and speed at the window level are identical during ascent and descent.
  • This symmetry helps streamline our calculations since the problems are often mirrored on both sides of the peak of the arc.
The ball's motion in the exercise illustrates how understanding the principles of projectile motion can simplify determining key variables such as time, speed, and ultimately, height.
Building Height Calculation
To find the height of the building, understanding the entire motion sequence of the steel ball is necessary. Here, both the details of the free-fall past the window and the return journey are crucial.

Calculate the distance the ball drops using the time and average velocity during the fall.
  • Remember that the ball's velocity as it passes the window gives insight into its speed just before and after falling the entire building height.
  • The calculation also involves using the formula \(h = \frac{1}{2}gt^2\), with the total time considered from the moment it leaves the rooftop until it stops on the ground below the window.
In the provided exercise, breaking the problem into sections assessing the fall time, window motion, and return enables the accurate calculation of the building height. This methodology showcases how physics principles assist in solving real-world problems.

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Most popular questions from this chapter

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