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Chapter 2: Problem 91

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

Short Answer

Expert verified
(a) 3.19 s (b) 1.33 s

Step by step solution

01

Understanding the Problem

We need to find the time it takes for a rock to fall two separate distances: the first 50 meters and the second 50 meters from a total height of 100 meters. The rock is falling under the influence of gravity, so we will apply the equations of motion for uniformly accelerated objects.
02

Write the Equation of Motion

The equation of motion to calculate the distance fallen under gravity is given by \[ s = ut + \frac{1}{2}gt^2 \]where:- \( s \) is the distance fallen,- \( u \) is the initial velocity,- \( g \) is the acceleration due to gravity (approximately \(9.8\, \mathrm{m/s^2}\)),- \( t \) is the time taken. For this problem, since the rock is initially at rest, \( u = 0 \).
03

Calculate Time for First 50 m

Using the equation \( s = \frac{1}{2}gt^2 \), we set \( s = 50 \) m and solve for \( t \):\[50 = \frac{1}{2} \times 9.8 \times t^2\]\[50 = 4.9t^2\]\[t^2 = 10.2\]\[t = \sqrt{10.2} \approx 3.19 \, \mathrm{s}\]Thus, it takes approximately 3.19 seconds to fall the first 50 meters.
04

Calculate Time for the Total 100 m

Now, calculate the total time to fall 100 meters using the same equation format:\[100 = \frac{1}{2} \times 9.8 \times T^2\]\[100 = 4.9T^2\]\[T^2 = 20.4\]\[T = \sqrt{20.4} \approx 4.52 \, \mathrm{s}\]The total time for 100 meters is approximately 4.52 seconds.
05

Calculate Time for Second 50 m

The time to fall the second 50 meters is the difference between the total time for 100 meters and the time for the first 50 meters:\[4.52\, \mathrm{s} - 3.19\, \mathrm{s} \approx 1.33 \, \mathrm{s}\]Thus, it takes approximately 1.33 seconds to fall the second 50 meters.

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Key Concepts

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

Understanding Free Fall
In physics, free fall occurs when the only force acting on an object is gravity. This means that any resistance, like air resistance, is neglected, allowing the object to accelerate downward purely under the influence of gravity. In a free-fall scenario, objects are initially at rest, meaning their starting velocity is zero. This aligns perfectly with motion equations where initial velocity is denoted as zero in calculations.

Let's consider why free fall is essential in our problem. When a rock falls off a cliff, it undergoes free fall, making it an ideal candidate for using motion equations.
  • No other forces affect the rock besides gravity.
  • We can determine the time taken to fall a certain distance easily.
Understanding these principles helps greatly with solving equations related to distance, velocity, and time when the object is under the influence of gravity.
The Role of Gravity
Gravity is the key player in free fall motion and directly influences how objects move when they are dropped. It is the acceleration that causes objects to increase their speed as they descend. On Earth, gravity accelerates all objects at approximately \(9.8 \, \mathrm{m/s^2}\).This constant acceleration means:
  • Objects will gain speed as they fall towards the ground.
  • This increment in speed continues until the object hits the ground or is acted upon by another force, like air resistance.
Gravity allows for the use of motion equations to predict how long it takes for an object to fall a given distance. The motion equations simplify calculations by considering gravity's constant acceleration, allowing students to solve problems about falling objects more easily.
Time Calculation in Free Fall
Calculating the time it takes for an object to fall a specific distance during free fall involves using motion equations where time, distance, and acceleration are connected. In these calculations, we focus on the equation \[ s = \frac{1}{2}gt^2 \].Here's a step-by-step breakdown of how the formula works:
  • \( s \) represents the distance the object has fallen, which is given in the problem.
  • \( g \) stands for the acceleration due to gravity, set as \(9.8 \, \mathrm{m/s^2}\).
  • \( t \) is the time, which is what we need to find.
By solving for \( t \), you discover how long it takes for the object to fall a specific distance.

In the given exercise, calculating the time for different distances can be done by first considering the total time it takes to reach a point and then the incremental difference to get from one point to another. This method effectively splits a larger free-fall scenario into smaller, more manageable parts for detailed analysis.

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Most popular questions from this chapter

In 1 km races, runner 1 on track 1 (with time \(2 \min .27 .95 \mathrm{~s}\) ) appears to be faster than runner 2 on track \(2(2 \min , 28.15 \mathrm{~s})\). Howcver, length \(L_{2}\) of track 2 might be slightly greater than length \(L_{1}\) of track 1. How large \(\operatorname{can} L_{2}-L_{1}\) be for us still to conclude that runner 1 is faster?

(a) If the position of a particle is given by \(x=20 t-5 t^{3}\), where \(x\) is in meters and \(t\) is in seconds, when, if ever, is the particle's velocity zero? (b) When is its acceleration \(a\) zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph \(x(t), v(t),\) and \(a(t)\)

A car can be braked to a stop from the autobahn-like speed of \(200 \mathrm{~km} / \mathrm{h}\) in \(170 \mathrm{~m}\). Assuming the acceleration is constant, Find its magnitude in (a) SI units and (b) in terms of g. (c) How much time \(T_{b}\) is required for the braking? Your reaction time \(T,\) is the time you require to perceive an emergency, move your foot to the brake. and begin the braking. If \(T_{r}=400 \mathrm{~ms}\), then (d) what is \(T_{b}\) in terms of \(T_{n}\) and \((\mathrm{c})\) is most of the full time required to stop spent in reacting or braking" Dark sunglasses delay the visual signals sent from the cyes to the visual cortex in the brain, increasing \(T_{r-}\) (f) In the extreme case in which \(T_{r}\) is increased by \(100 \mathrm{~ms}\), how much farther does the car travel during your reaction time?

A certain juggler usually tosses halls vertically to a height \(H\). To what height must they be tossed if they are to spend twice as much time in the air?

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 spced docs it hit the ground?
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