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A bolt is dropped from a bridge under construction, falling \(90 \mathrm{~m}\) to the valley below the bridge. (a) In how much time does it pass through the last \(20 \%\) of its fall? What is its speed (b) when it begins that last \(20 \%\) of its fall and \((\mathrm{c})\) when it reaches the valley beneath the bridge?

Short Answer

Expert verified
(a) 0.46 s; (b) 37.5 m/s; (c) 42.0 m/s.

Step by step solution

01

Analyze the Problem

The bolt is dropped from a height of 90 m. We need to calculate the time and speed at certain points of its fall. The problem involves two main concepts: free fall motion and velocity calculations.
02

Calculate Total Fall Time

We need to find the total time taken for the bolt to fall 90 m. Using the equation for free fall, \[ s = \frac{1}{2} g t^2 \] where \(s\) is the distance fallen (90 m) and \(g\) is the acceleration due to gravity (approximately \(9.8\, \text{m/s}^2\)), we get:\[ 90 = \frac{1}{2} \times 9.8 \times t^2 \]Solving for \(t\), we find:\[ t^2 = \frac{180}{9.8} \approx 18.37 \Rightarrow t \approx \sqrt{18.37} \approx 4.29 \text{ seconds} \]
03

Determine Last 20% of Fall Distance

The last 20% of the fall distance is \(0.2 \times 90 = 18\, \text{m}\). Thus, the bolt falls the first \(90 - 18 = 72\, \text{m}\) before beginning this final part of the fall.
04

Calculate Time to Reach Last 20%

Now, calculate the time it takes to fall the first 72 m using the same free-fall equation:\[ 72 = \frac{1}{2} \times 9.8 \times t_{72}^2 \]Solving for \(t_{72}\), we find:\[ t_{72}^2 = \frac{144}{9.8} \approx 14.69 \Rightarrow t_{72} \approx \sqrt{14.69} \approx 3.83 \text{ seconds} \]
05

Calculate Time for Last 20% of Fall

The time to fall the last 20% is the difference between the total time and the time to reach 72 m:\[ t_{last 20} = 4.29 - 3.83 = 0.46\, \text{seconds} \]
06

Speed at Beginning of Last 20%

Calculate the speed of the bolt when it reaches 72 m using:\[ v = g \times t_{72} \]\[ v = 9.8 \times 3.83 \approx 37.5\, \text{m/s} \]
07

Speed Upon Reaching the Valley

The final speed when reaching the valley is:\[ v_{final} = g \times t_{total} \]\[ v_{final} = 9.8 \times 4.29 \approx 42.0\, \text{m/s} \]

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Key Concepts

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

Velocity Calculations
Velocity is a crucial factor in understanding the behavior of objects in motion, especially in free fall. When an object falls under the influence of gravity, its velocity changes as it descends. To compute the velocity, you need to know the time the object has been falling and the acceleration it experiences during the fall.

In free fall, the velocity can be calculated using the formula: \( v = g \times t \), where \( v \) stands for velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time. This formula shows that velocity increases linearly over time while falling under gravity alone.
  • During the initial seconds of free fall, the speed is relatively low.
  • As time progresses, the speed continues to increase at a steady rate determined by gravity.
This constant increment explains why objects in free fall reach higher speeds over time.
Acceleration Due to Gravity
The acceleration due to gravity is a pivotal element in motion studies, particularly in free fall scenarios. On Earth, this acceleration, symbolized as \( g \), is approximately \( 9.8 \, \text{m/s}^2 \). This constant value indicates how fast the velocity of a falling object increases and is the same for all objects, regardless of their mass or size, when air resistance is negligible.

To grasp how acceleration due to gravity functions, consider:
  • Each second a free-fall object drops, it's gaining around 9.8 meters per second in speed.
  • This means if you start at 0 m/s, after 1 second, you are moving at 9.8 m/s; after 2 seconds, at approximately 19.6 m/s, and so on.
Such predictability makes it easy to calculate the behavior of falling objects in theoretical settings.
Distance-Time Relationship
Understanding the distance-time relationship in free fall is fundamental to solving problems like the one provided. During free fall, the distance an object falls is directly related to both time and acceleration due to gravity.

The formula \( s = \frac{1}{2} g t^2 \) encapsulates this relationship, where \( s \) is the distance fallen, \( g \) is gravitational acceleration, and \( t \) is time. The equation reflects how the distance increases quadratically with time, highlighting that the longer an object falls, the faster it accumulates distance.
  • Initially, the object covers a small distance in the first few seconds.
  • As the fall continues, the distance covered grows rapidly due to the square of time in the equation.
  • For example, doubling the time doesn't just double the distance—it quadruples it.
Recognizing this quadratic relationship helps predict how an object behaves under gravity alone.
Projectile Motion
Projectile motion involves objects that are thrown or projected into the air and subject only to gravity as they move along a curved path. Key to understanding projectile motion is recognizing the independence of horizontal and vertical motions.

In the specific case of a free-falling object like a bolt dropped vertically with no initial horizontal velocity, the interesting part is the vertical motion driven solely by gravity.
  • The only force here is gravity, pulling the object downward.
  • Despite the absence of initial horizontal motion, the principles of projectile motion still apply in analysis.
  • Vertical component analysis allows for precise calculation of how fast an object will be moving at any point in time.
Understanding projectile motion, even in simple vertical free fall, provides insight into how objects behave when influenced primarily by gravity, free of other external forces.

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Most popular questions from this chapter

A pilot flies horizontally at \(1300 \mathrm{~km} / \mathrm{h}\), at height \(h=35 \mathrm{~m}\) above initially level ground. However, at time \(t=0\), the pilot begins to fly over ground sloping upward at angle \(\theta=4.3^{\circ}\) (Fig. \(2-41\) ). If the pilot does not change the airplane's heading, at what time \(t\) does the plane strike the ground?

A ball of moist clay falls \(15.0 \mathrm{~m}\) to the ground. It is in contact with the ground for \(20.0 \mathrm{~ms}\) before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?

In an arcade video game, a spot is programmed to move across the screen according to \(x=9.00 t-0.750 t^{3}\), where \(x\) is distance in centimeters measured from the left edge of the screen and \(t\) is time in seconds. When the spot reaches a screen edge, at either \(x=0\) or \(x=15.0 \mathrm{~cm}, t\) is reset to 0 and the spot starts moving again according to \(x(t) .\) (a) At what time after starting is the spot instantaneously at rest? (b) At what value of \(x\) does this occur? (c) What is the spot's acceleration (including sign) when this occurs? (d) Is it moving right or left just prior to coming to rest? (e) Just after? (f) At what time \(t>0\) does it first reach an edge of the screen?

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

A basketball player grabbing a rebound jumps \(76.0 \mathrm{~cm}\) vertically. How much total time (ascent and descent) does the player spend (a) in the top \(15.0 \mathrm{~cm}\) of this jump and (b) in the bottom \(15.0 \mathrm{~cm} ?\) (The player seems to hang in the air at the top.)

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