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Although the altitude may vary considerably, hailstones sometimes originate around \(500 \mathrm{~m}\) (about \(1500 \mathrm{ft}\) ) above the ground. (a) If we ignore air drag, how fast will these hailstones be moving when they reach the ground, assuming that they start from rest? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{mph} .\) (b) From your own experience, are hailstones actually falling that fast when they reach the ground? Why not? What happens to most of the initial potential energy?

Short Answer

Expert verified
(a) 99 m/s or 221 mph; (b) No, because air drag slows them down.

Step by step solution

01

Define the problem

We need to find the final velocity of hailstones falling from a height of 500 m, assuming no air resistance. We will use the equations of motion for free falling objects.
02

Apply the equation of motion

Using the equation \( v^2 = u^2 + 2gh \), where \( u = 0 \) m/s is the initial velocity, \( g = 9.8 \) m/s² is the acceleration due to gravity, and \( h = 500 \) m is the height, we find:\[ v^2 = 0 + 2 \times 9.8 \times 500 \]
03

Calculate the final velocity in m/s

Solve for \( v \):\[ v^2 = 9800 \]\[ v = \sqrt{9800} \approx 99 \text{ m/s} \]
04

Convert m/s to mph

Use the conversion factor 1 m/s = 2.237 mph:\[ v \approx 99 \times 2.237 \approx 221 \text{ mph} \]
05

Consider real-world factors

In reality, hailstones do not fall at 99 m/s (221 mph), primarily because of air resistance, which acts against gravity and reduces their velocity. As a result, most of the potential energy of the hailstones is converted into heat and turbulent air currents.

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Key Concepts

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

Hailstones Velocity
When hailstones fall from a height, their speed increases progressively until they reach the ground. This speed is known as their final velocity. In an ideal scenario, where air resistance is ignored, the only force acting on the hailstones is gravity. This situation is termed as 'free fall'. The velocity of these hailstones is determined using the equations of motion, assuming they start from rest.
In our exercise, the hailstones begin falling from a height of 500 meters. A key equation to determine their speed when they hit the ground is:
  • \( v^2 = u^2 + 2gh \)
  • Where \( v \) is the final velocity, \( u \) is the initial velocity (0 m/s for free fall), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height (500 m).
By entering these values, we derive that the final speed, ignoring air resistance, would be approximately 99 m/s, or about 221 mph when converted. This illustrates how quickly hailstones can potentially travel if other forces are not taken into account.
Air Resistance Effects
In reality, hailstones do not accelerate solely under the influence of gravity. Air resistance, also known as drag force, plays a significant role in slowing their fall. This is why actual falling speeds are often much lower than theoretical calculations in a vacuum.
The drag force increases with velocity and acts in the opposite direction to the motion of hailstones. Essentially, air resistance reduces the effectiveness of gravity, limiting the terminal velocity - the constant speed achieved by the hail as it falls through the air.
This interaction is why most hailstones don't reach the high speeds predicted without drag. Instead, they descend at speeds often around 20 to 40 m/s, well below the 99 m/s figure we've calculated by ignoring air resistance. Moreover, the kinetic energy gets transformed into heat due to collisions with air particles, resulting in less energy available for further acceleration.
Potential Energy Conversion
As hailstones fall and gain speed, they are converting their potential energy into kinetic energy. Potential energy is the energy possessed by an object due to its height above the ground. For hailstones at a height of 500 meters, a significant amount of such energy is present initially.
The formula for potential energy is:
  • \( PE = mgh \)
  • Where \( PE \) is potential energy, \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.
As the hailstones descend, this potential energy is transformed into kinetic energy (motion energy), following the conservation of energy principle that energy in a closed system remains constant.
However, not all potential energy converts into kinetic energy due to air resistance. Instead, some of it gets converted into other forms like heat and sound as the hailstone interacts with air molecules. This conversion explains why hailstones move slower in reality and do not hit the ground with predicted high speeds.
Equations of Motion
Equations of motion are vital in predicting how objects move under different forces. They help calculate motion properties like velocity, displacement, and acceleration. In the case of hailstones, these equations allow us to understand their velocity as they fall under the influence of gravity.
The primary equation used here is:
  • \( v^2 = u^2 + 2gh \)
  • It connects the initial velocity \( u \), final velocity \( v \), acceleration due to gravity \( g \), and height \( h \).
This particular formula is part of the set of kinematic equations, applicable for constant acceleration scenarios. By setting the initial velocity \( u \) to zero for free-fall, we can calculate the final speed of free-falling objects.
While they provide an ideal calculation framework, remember to factor in environmental forces like drag for real-world applications, affirming their importance in physics education.

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