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Chapter 2: Problem 91

A rock is dropped from a 100 -m-high cliff. How long does it take to fall (a) the first \(50 \mathrm{~m}\) and \((\mathrm{b})\) the second \(50 \mathrm{~m}\) ?

Short Answer

Expert verified
It takes approximately 3.19 s for the first 50 m drop and an additional 1.95 s for the next 50 m.

Step by step solution

01

Understand the Problem

We need to determine the time taken for a rock to fall two different segments of a 100-meter drop: the first 50 meters and the second 50 meters.
02

Identify Known Variables

The initial position, \( s_0 \), is 100m for the entire drop. For the first 50m drop, the final position, \( s \), is 50m. For the second 50m drop, the final position is 0m. The acceleration, \( a \), due to gravity is \( 9.8 \text{ m/s}^2 \). Initial velocity \( v_0 \) is 0 m/s since the rock is dropped.
03

First 50 m Drop Time Calculation

Use the equation of motion: \[ s = s_0 + v_0 t + \frac{1}{2} a t^2 \] Substitute \( s_0 = 100 \), \( s = 50 \), \( a = 9.8 \), and \( v_0 = 0 \): \[ 50 = 100 + 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \] Simplify to solve for \( t \): \[ 0.5 \cdot 9.8 \cdot t^2 = 50 \] \[ 4.9 t^2 = 50 \] \[ t^2 = \frac{50}{4.9} \] \[ t = \sqrt{\frac{50}{4.9}} \approx 3.19 \text{ s} \]
04

Second 50 m Drop Time Calculation

For the second 50 m drop, the rock has already fallen for 3.19 s, reaching 50 m above the ground. To find the total time to fall to 0 meters, use the initial velocity of this segment as \( v = 0 + 9.8 \cdot 3.19 = 31.262 \text{ m/s} \). Now use: \[ s = s_0 + v_0 t + \frac{1}{2} a t^2 \] Here \( s_0 = 50 \), \( s = 0 \), \( a = 9.8 \), and \( v_0 = 31.262 \): \[ 0 = 50 + 31.262 t + \frac{1}{2} \cdot 9.8 \cdot t^2 \] Rearrange to solve for \( t \) using the quadratic formula: \[ -4.9 t^2 - 31.262 t + 50 = 0 \] Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -4.9 \), \( b = -31.262 \), \( c = 50 \): \[ t = \frac{31.262 \pm \sqrt{31.262^2 - 4 \cdot -4.9 \cdot 50}}{2 \cdot -4.9} \approx 1.949 \text{ s} \] after validating relevant solution.

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Key Concepts

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

Equations of Motion
Equations of motion help us predict how objects move based on certain forces. In our exercise, these equations simplify the analysis of the motion of the rock falling under gravity. In the context of free fall, there are a few key points and formulas to remember:
  • The rock starts from rest, so the initial velocity, \( v_0 \), is zero.
  • The main equation of motion we're using is: \[ s = s_0 + v_0 t + \frac{1}{2} a t^2 \]
  • This equation calculates displacement \( s \) over time \( t \), with initial position \( s_0 \), velocity \( v_0 \), and acceleration \( a \).
  • In free fall, the only force at work is gravity, affecting the object constantly and uniformly.
For both parts of the exercise, knowing initial conditions (position and velocity) allows us to solve for time, \( t \), using the motion equation. Understanding these motions helps us apply the right math to find our answers.
Acceleration due to Gravity
Acceleration due to gravity is a central character in the analysis of any free-fall scenario. On Earth, this effect is pretty constant. It pulls objects toward the surface at a rate of \( 9.8 \text{ m/s}^2 \). This means, if you were to drop a rock from a height, every second, its velocity increases by \( 9.8 \text{ m/s} \).
Understanding this concept makes analyzing the rock's fall more predictable. You can anticipate how fast and how far the rock travels over time, knowing gravity's consist growth effect. Also, recognizing that the acceleration doesn't change means we can confidently apply the same formulas across different objects and conditions on Earth.
  • It's important to remember that gravity solely influences the object in free-fall by increasing its speed.
  • When solving problems, use the consistent value of \( 9.8 \text{ m/s}^2 \) to simplify calculations.
  • This regularity helps us utilize equations of motion effectively in a sequence without complicating form change.
Quadratic Formula
The quadratic formula is essential in solving equations where variable powers beyond one (linear) appear. In our case, it's crucial for handling the second part of the exercise where the rock's journey culminates in ground-surfacing. Normally, the quadratic equation exhibits this form: \[ ax^2 + bx + c = 0 \]
For the rock's fall, we need solutions for time, \( t \), translating to finding roots of this kind of equation. Here's how the quadratic formula is structured:
  • \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This formula comes in handy when looking at polynomial equations where the highest degree is two. It provides one or two solutions helping us discover times when the rock hits certain points during its descent.
For the second segment, we recognize these coefficients from our executed formula for position (substituting known values of beginning speed, acceleration, and distance-related initial position). From our setting: \( a = -4.9 \), \( b = -31.262 \), and \( c = 50 \). Unveiling its roots gets us our elapsed time effectively.
Grasping the quadratic solution methodology helps build confidence in tackling similar physics problems with complex motion equations.

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Most popular questions from this chapter

A car moving with constant acceleration covered the distance between two points \(60.0 \mathrm{~m}\) apart in \(6.00 \mathrm{~s}\). Its speed as it passed the second point was \(15.0 \mathrm{~m} / \mathrm{s}\). (a) What was the speed at the first point? (b) What was the magnitude of the acceleration? (c) At what prior distance from the first point was the car at rest? (d) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the car, from rest \((t=0)\).

The position function \(x(t)\) of a particle moving along an \(x\) axis is \(x=4.0-6.0 t^{2}\), with \(x\) in meters and \(t\) in seconds. (a) At what time and (b) where does the particle (momentarily) stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph \(x\) versus \(t\) for the range \(-5 \mathrm{~s}\) to \(+5 \mathrm{~s}\). (f) To shift the curve rightward on the graph, should we include the term \(+20 t\) or the term \(-20 t\) in \(x(t) ?(\mathrm{~g})\) Does that inclusion increase or decrease the value of \(x\) at which the particle momentarily stops?

To set a speed record in a measured (straight-line) distance \(d\), a race car must be driven first in one direction (in time \(t_{1}\) ) and then in the opposite direction (in time \(t_{2}\) ). (a) To eliminate the effects of the wind and obtain the car's speed \(v_{c}\) in a windless situation, should we find the average of \(d / t_{1}\) and \(d / t_{2}\) (method 1) or should we divide \(d\) by the average of \(t_{1}\) and \(t_{2} ?\) (b) What is the fractional difference in the two methods when a steady wind blows along the car's route and the ratio of the wind speed \(v_{w}\) to the car's speed \(v_{c}\) is \(0.0240 ?\)

An electron has a constant acceleration of \(+3.2 \mathrm{~m} / \mathrm{s}^{2}\). At a certain instant its velocity is \(+9.6 \mathrm{~m} / \mathrm{s}\). What is its velocity (a) \(2.5 \mathrm{~s}\) earlier and (b) \(2.5\) s later?

The position of an object moving along an \(x\) axis is given by \(x=3 t-4 t^{2}+t^{3}\), where \(x\) is in meters and \(t\) in seconds. Find the position of the object at the following values of \(t:\) (a) \(1 \mathrm{~s}\), (b) \(2 \mathrm{~s}\), (c) \(3 \mathrm{~s}\), and \((\mathrm{d}) 4 \mathrm{~s}\). (e) What is the object's displacement between \(t=0\) and \(t=4 \mathrm{~s} ?\) (f) What is its average velocity for the time interval from \(t=2 \mathrm{~s}\) to \(t=4 \mathrm{~s} ?(\mathrm{~g})\) Graph \(x\) versus \(t\) for \(0 \leq t \leq 4 \mathrm{~s}\) and indicate how the answer for (f) can be found on the graph.
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