/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 A raindrop falls from a cloud \(... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 104

A raindrop falls from a cloud \(100 \mathrm{m}\) above the ground. Neglect air resistance. What is the speed of the raindrop when it hits the ground? Is this a reasonable number?

Short Answer

Expert verified
The speed of the raindrop when it hits the ground, neglecting air resistance, is calculated as \(44.3 ~\mathrm{m/s}\) (approx. \(160~\mathrm{km/h}\)). This value might not be reasonable in reality as air resistance would significantly lower the actual speed. However, under the assumption of no air resistance, the calculation is correct.

Step by step solution

01

Identify the variables

Here is the given information: - Initial height (distance) from the cloud to the ground: \(h = 100 ~\mathrm{m}\) - Acceleration due to gravity: \(g = 9.81 ~\mathrm{m/s^2}\) Our goal is to find the final velocity (\(v\)) of the raindrop when it hits the ground.
02

Choose a kinematic equation

A suitable kinematic equation to use in this situation is: \(v^2 = u^2 + 2as\) where: - \(v\) is the final velocity of the raindrop, - \(u\) is the initial velocity of the raindrop (which is 0, since it starts from rest), - \(a\) is the acceleration (in this case, due to gravity), - \(s\) is the distance the raindrop travels.
03

Substitute the values into the equation

Plug in the given values (with \(u = 0\), \(a = g\), and \(s = h\)): \(v^2 = 0^2 + 2 \times 9.81 ~\mathrm{m/s^2} \times 100~\mathrm{m}\)
04

Calculate the final velocity

Now we can calculate the final velocity of the raindrop: \(v^2 = 2 \times 9.81~\mathrm{m/s^2} \times 100~\mathrm{m} = 1962~\mathrm{m^2/s^2}\) To find the final velocity, we'll take the square root of the result: \(v = \sqrt{1962~\mathrm{m^2/s^2}} = 44.3~\mathrm{m/s}\)
05

Evaluate the result

The calculated speed of the raindrop when it hits the ground is \(44.3 ~\mathrm{m/s}\), or approximately \(160~\mathrm{km/h}\). In reality, air resistance would play a significant role in slowing down the raindrop, so the actual speed would be lower than this calculated value. Considering this, the answer we got might not be reasonable if we account for air resistance. However, under the assumption of no air resistance, the calculation is correct.

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Key Concepts

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

Free Fall Motion
When an object is subject only to the force of gravity, it is said to be in free fall motion. This means that the only force acting on the object is gravity, which causes it to accelerate downwards. Importantly, air resistance in this context is ignored, allowing us to treat the acceleration as constant, which greatly simplifies our calculations.

In the case of the raindrop from our exercise, it starts from rest and accelerates due to gravity, without any other forces like air resistance to impede its motion. This assumption is key for using the kinematic equations accurately and obtaining a theoretical value for the final velocity of the raindrop as it hits the ground.
Acceleration Due to Gravity
The acceleration due to gravity, denoted as \(g\), is a constant value that represents how quickly an object speeds up when it is in free fall near the Earth's surface. This constant is roughly \(9.81 \mathrm{m/s^2}\) on Earth, which means that every second, the velocity of an object in free fall will increase by about \(9.81 \mathrm{m/s}\) if air resistance is not considered.

Understanding this concept is essential because it provides us with the \(a\) value in our kinematic equations necessary to solve for the final velocity. Whether it鈥檚 a raindrop or another object dropped from a height, this acceleration remains consistent and is crucial in predicting the object鈥檚 behavior as it falls.
Final Velocity Calculation
Calculating the final velocity of an object in free fall involves the use of kinematic equations that describe motion with constant acceleration. The equation \(v^2 = u^2 + 2as\) is specifically useful for situations where we want to find the final velocity \(v\) of an object starting from an initial velocity \(u\) after accelerating over a distance \(s\).

In our exercise, we started with the raindrop at rest, which means \(u = 0\). The distance \(s\) is the height from which the raindrop fell, and \(a\) is the acceleration due to gravity, \(g\). By substituting these values into our chosen kinematic equation, we were able to determine the theoretical speed of the raindrop upon impact with the ground. It's important to note that this calculation assumes no air resistance, hence the result can only be considered accurate under ideal conditions.

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Most popular questions from this chapter

An airplane accelerates at \(5.0 \mathrm{m} / \mathrm{s}^{2}\) for \(30.0 \mathrm{s}\). During this time, it covers a distance of \(10.0 \mathrm{km}\). What are the initial and final velocities of the airplane?

A particle has a constant acceleration of \(6.0 \mathrm{m} / \mathrm{s}^{2}\) (a) If its initial velocity is \(2.0 \mathrm{m} / \mathrm{s}\), at what time is its displacement 5.0 m? (b) What is its velocity at that time?

(a) A light-rail commuter train accelerates at a rate of \(1.35 \mathrm{m} / \mathrm{s}^{2} .\) How long does it take to reach its top speed of 80.0 \(\mathrm{km} / \mathrm{h}\), starting from rest? (b) The same train ordinarily decelerates at a rate of \(1.65 \mathrm{m} / \mathrm{s}^{2} .\) How long does it take to come to a stop from its top speed? (c) In emergencies, the train can decelerate more rapidly, coming to rest from \(80.0 \mathrm{km} / \mathrm{h}\) in \(8.30 \mathrm{s}\). What is its emergency acceleration in meters per second squared?

On February 15, 2013, a superbolide meteor (brighter Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity of the blast wave? b) Compare this with the speed of sound, which is \(343 \mathrm{m} / \mathrm{s}\) at sea level.

Assume an intercontinental ballistic missile goes from rest to a suborbital speed of \(6.50 \mathrm{km} / \mathrm{s}\) in \(60.0 \mathrm{s}\) (the actual speed and time are classified). What is its average acceleration in meters per second and in multiples of \(\bar{g}\) \(\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) ?\)
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