Problem 44 - College Physics

Question

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

Step-by-Step Solution

Answer

(a) 99 m/s or 221.4 mph; (b) No, because air drag reduces their speed.

Step 1: Identify the Known Variables

We know that the hailstones originate from an initial height, \( h = 500 \, \text{m} \), and they start from rest, which means their initial velocity, \( v_0 = 0 \, \text{m/s} \). The acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \).

Step 2: Apply Energy Conservation

Neglecting air resistance, we will use the conservation of mechanical energy. The potential energy at height \( h \) is converted entirely into kinetic energy when the hailstones reach the ground. Thus, \( mgh = \frac{1}{2}mv^2 \), where \( v \) is the speed upon reaching the ground.

Step 3: Solve for Final Velocity

Since \( m \) appears on both sides of the equation, it cancels out. Thus, \( gh = \frac{1}{2}v^2 \). Solving for \( v \), we get \( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 500 \, \text{m}} \).

Step 4: Calculate Final Velocity in m/s

Plug the values into the equation: \( v = \sqrt{2 \times 9.8 \times 500} = \sqrt{9800} \approx 99 \, \text{m/s} \).

Step 5: Convert Velocity to mph

To convert \( 99 \, \text{m/s} \) to mph, use the conversion factor \( 1 \, \text{m/s} = 2.237 \, \text{mph} \). Thus, \( 99 \, \text{m/s} \times 2.237 \approx 221.4 \, \text{mph} \).

Step 6: Consider Real-World Factors

In reality, hailstones do not reach these speeds because air resistance (drag) significantly slows them down. Air resistance converts much of the potential energy into heat or dissipates it due to the drag force.

Key Concepts

Kinematic Equations

Kinematic equations are essential tools in physics to describe the motion of objects. When working with problems involving motion under constant acceleration, such as the fall of a hailstone due to gravity, these equations become very useful. In this exercise, we focus on finding the final velocity of the hailstone just before it hits the ground.

Here, we're assuming an initial velocity of zero since the hailstone starts from rest and a constant acceleration due to gravity which is approximately 9.8 m/s² near the surface of the Earth. Using the kinematic equation:

• \( v^2 = v_0^2 + 2a d \)

Where:

• \( v_0 \) is the initial velocity
• \( a \) is the acceleration due to gravity
  
• \( d \) is the distance or height
  

We compute the final velocity \( v \). It's a straightforward computation if you methodically identify and insert the given values. This equation helps you predict the motion of an object based solely on its initial state without needing to observe every moment of its journey.

Energy Conservation

Energy conservation is a core principle in physics, stating that energy in a closed system remains constant. In this exercise, we assume no air resistance, thus energy conservation aids in solving for the final velocity of the hailstone as it transforms potential energy at a height into kinetic energy.

The potential energy (PE) at the starting point is given by:

• \( ext{PE} = mgh \)

Where:

• \( m \) is the mass of the hailstone,
• \( g \), the gravitational acceleration,
• \( h \), the height from which it falls.

Kinetic energy (KE) upon reaching the ground is:

• \( ext{KE} = \frac{1}{2}mv^2 \)

Since energy is conserved:

• \( mgh = \frac{1}{2}mv^2 \)
• Solving for \( v \), the mass \( m \) cancels.

This clear transformation of energy offers a streamlined way to deduce physical results, granting insight into how potential energy translates into kinetic energy during the fall.

Impact of Air Resistance

Air resistance, often referred to as drag, significantly alters the motion of falling objects like hailstones. Although theory calculations show velocities up to 99 m/s, in the real world, hailstones never fall this fast. This deceleration occurs because air slows the fall.

The force of air resistance builds as the hailstone accelerates, reaching a point where it equals the gravitational pull. This equilibrium halts further acceleration, causing the hailstone to fall at a terminal velocity instead. The terminal velocity is notably slower than the theoretical speed calculated without air resistance.

Consequently, while potential energy should convert into kinetic energy without any loss, the presence of drag shifts some of this energy into other forms, such as heat due to friction. This real-world consideration is vital in accurately understanding how energy and force dynamics influence object speeds in practical scenarios.