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Chapter 2: Problem 88

Cliff Height. You are climbing in the High Sierra where you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top and 10.0 s later hear the sound of it hitting the ground at the foot of the cliff. (a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 \(\mathrm{m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.

Short Answer

Expert verified
(a) The height of the cliff is approximately 384.5 meters. (b) You would overestimate the height if ignoring sound's travel time.

Step by step solution

01

Determine the Total Time

The total time given is 10 seconds, which is the sum of the time taken by the rock to fall and the time taken for the sound to travel back to you.
02

Set Up Equations for Rock and Sound

Let the time taken for the rock to fall be \( t_1 \) and the time taken for sound to travel back be \( t_2 \). We have:\[ t_1 + t_2 = 10 \] Let the height of the cliff be \( h \). For the rock:\[ h = \frac{1}{2} g t_1^2 \] (where \( g = 9.8 \, \text{meters per second squared} \)).And for the sound: \[ h = 330 t_2 \] (since speed of sound = 330 m/s).
03

Solve the Equations

Substitute \( t_2 = 10 - t_1 \) into the equation for sound:\[ h = 330(10 - t_1) \]Set equal the two equations for \( h \):\[ \frac{1}{2} g t_1^2 = 330(10 - t_1) \]Simplify and solve for \( t_1 \):\[ 4.9 t_1^2 = 3300 - 330 t_1 \]Rearrange to find:\[ 4.9 t_1^2 + 330 t_1 - 3300 = 0 \]Use the quadratic formula \( t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( t_1 \).
04

Apply the Quadratic Formula

Here, \( a = 4.9 \), \( b = 330 \), and \( c = -3300 \). Plug these values into the quadratic formula:\[ t_1 = \frac{-330 \pm \sqrt{330^2 - 4 \times 4.9 \times (-3300)}}{2 \times 4.9} \]Calculate:\[ t_1 = \frac{-330 \pm \sqrt{108900 + 64680}}{9.8} \]\[ t_1 = \frac{-330 \pm \sqrt{173580}}{9.8} \]\[ t_1 = \frac{-330 \pm 416.57}{9.8} \]Choose the positive root, since time cannot be negative:\( t_1 \approx 8.85 \, \text{seconds} \)
05

Calculate the Height of the Cliff

Substitute \( t_1 \approx 8.85 \) back into the falling rock equation:\[ h = \frac{1}{2} \times 9.8 \times (8.85)^2 \]Calculate the height:\[ h \approx 384.5 \, \text{meters} \]
06

Consider Ignoring Sound's Travel Time

If you had ignored the time required for sound to travel back, you would have used 10 seconds as the time for the rock to fall. This overestimates the falling time, leading to a larger calculation for height, thus overestimating the height of the cliff.

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Key Concepts

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

Free Fall Calculations
Free fall is one of the most fundamental concepts in physics, often serving as a basis for more complex motion studies. When we talk about free fall, we are assuming that the only force acting on an object is gravity. Under such circumstances, the acceleration due to gravity is denoted as \(g\) and has a standard value of approximately \(9.8 \, \mathrm{m/s^2}\).

When an object is in free fall, it starts from rest (unless stated otherwise), and its motion involves increasing speed as it descends. The basic equation for free fall can be represented by:
  • \( h = \frac{1}{2} g t^2 \)
where:
  • \(h\) is the height or distance fallen.
  • \(g\) is the acceleration due to gravity.
  • \(t\) is the time taken to fall.
In our cliff problem, we find the time \(t_1\) taken by the rock to fall as part of our solution to calculate the height \(h\) of the cliff using this equation, thereby isolating any other factors like air resistance that might affect the descent.
Sound Velocity
The speed or velocity of sound is another crucial concept in our physics problem. Sound travels through air at a velocity that is often approximated to \(330 \, \mathrm{m/s}\). However, this speed can be influenced by several factors, including temperature, humidity, and altitude. For simplicity in our exercise, we use the standard speed of sound for calculations.

In the cliff example, sound velocity helps us calculate the time \(t_2\) it takes for the sound to travel from the point of impact at the base of the cliff back up to the point where it was heard at the top. This aspect is critical because the total time of \(10\) seconds includes both the falling time and the time for the sound to travel back up. The equation used is:
  • \( h = 330 t_2 \)
Here, \(h\) is the height of the cliff and \(t_2\) the time taken by sound to reach the observer, allowing us to split the total time into measurable components.
Quadratic Equations in Physics
Quadratic equations often appear in physics when dealing with motion and forces, particularly when two forces or motions need to be balanced, such as in our free fall problem. The motion of the rock falling and the sound traveling can be represented as quadratics.

In the given problem, after establishing the expressions for both the free fall and sound travel, we end up with a quadratic equation:
  • \( 4.9 t_1^2 + 330 t_1 - 3300 = 0 \)
To solve this equation, the quadratic formula is employed:
  • \( t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This formula helps find the values of \(t_1\), leading to a solution where we choose the positive root as time cannot be negative. This quadratic solution is essential to determining the correct falling time, which is then applied to find the height using previously discussed free fall equations.

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Most popular questions from this chapter

An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of \(2.10 \mathrm{m} / \mathrm{s}^{2},\) and the antomobile an acceleration of 3.40 \(\mathrm{m} / \mathrm{s}^{2} .\) The automobile overtakes the truck after the truck has moved 40.0 \(\mathrm{m}\) . (a) How much time does it take the automobile to overtake the truck? (b) How far was the automobile behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x=0\) at the initial location of the truck.

A car's velocity as a function of time is given by \(v_{x}(t)=\) \(\alpha+\beta t^{2},\) where \(\alpha=3.00 \mathrm{m} / \mathrm{s}\) and \(\beta=0.100 \mathrm{m} / \mathrm{s}^{3}\) , (a) Calculate the average acceleration for the time interval \(t=0\) to \(t=5.00 \mathrm{s}\) (b) Calculate the instantaneous acceleration for \(t=0\) and \(t=\) 5.00 \(\mathrm{s}\) . (c) Draw accurate \(v_{x}-t\) and \(a_{x}-t\) graphs for the car's motion between \(t=0\) and \(t=5.00 \mathrm{s} .\)

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

A hot-air balloonist, rising vertically with a constant velocity of magnitude \(5.00 \mathrm{m} / \mathrm{s},\) releases a sandbag at an instant when the balloon is 40.0 \(\mathrm{m}\) above the ground (Fig. 2.41\()\) . After it is released, the sandbag is in frec fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds s strike after its release will the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_{y}-t\) \(v_{y^{-} t},\) and \(y-t\) graphs for the motion.

Mars Landing. In Jamuary 2004 , NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages: Stage A: Friction with the atmosphere reduced the speed from \(19,300 \mathrm{km} / \mathrm{h}\) to 1600 \(\mathrm{km} / \mathrm{h}\) in 4.0 \(\mathrm{min.}\) Stage \(\mathrm{B} : \mathrm{A}\) parachute then opened to slow it down to 321 \(\mathrm{km} / \mathrm{h}\) in 94 \(\mathrm{s}\) . Stage \(C :\) Retro rockets then fired to reduce its speed to zero over a distance of 75 \(\mathrm{m} .\) Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant. (a) Find the rocket's accoleration (in \(\mathrm{m} / \mathrm{s}^{2} )\) during cach stage. (b) What total distance \((\text { in } \mathrm{km})\) did the rocket travel during stages \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) ?
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