Answer :
a)The speed of the raindrop falling in the absence of air drag is approximately 7.659 × 10³ m/s.
b)The speed of the raindrop at the end of 3700 m when there is air drag is approximately 8.136 m/s.
(a) To calculate the speed of the raindrop falling in the absence of air drag, we can use the principle of conservation of mechanical energy. The potential energy lost by the raindrop as it falls is converted into kinetic energy.
The potential energy lost by the raindrop is given by:
Potential energy = m * g * h
where
m = mass of the raindrop
g = acceleration due to gravity (approximately 9.8 m/s²)
h = vertical distance fallen (3700 m)
The mass of the raindrop can be calculated using its volume and density:
Volume of a sphere = (4/3) * π * r³
where r = radius of the raindrop = 3.5 mm / 2 = 1.75 mm = 0.00175 m
Mass of the raindrop = Volume * Density
Substituting the values into the formula, we have:
Mass of the raindrop = [(4/3) * π * (0.00175 m)³] * 1000 kg/m³
Calculating the mass, we find:
Mass of the raindrop ≈ 6.614 × 10⁻⁸ kg
Now, we can calculate the potential energy:
Potential energy = (6.614 × 10⁻⁸ kg) * (9.8 m/s²) * (3700 m)
Calculating the potential energy, we find:
Potential energy ≈ 2.349 J
Since the potential energy is converted entirely into kinetic energy, we can equate them:
Potential energy = Kinetic energy
2.349 J = (1/2) * m * v²
Solving for v (the speed), we have:
v = √[(2 * 2.349 J) / m]
Substituting the values, we get:
v ≈ √[(2 * 2.349 J) / (6.614 × 10⁻⁸ kg)]
Calculating the speed, we find:
v ≈ 7.659 × 10³ m/s
Therefore, the speed of the raindrop falling in the absence of air drag is approximately 7.659 × 10³ m/s.
(b) To calculate the speed of the raindrop at the end of 3700 m when there is air drag, we need to consider the opposing force of air resistance.
When the raindrop falls, it accelerates due to the force of gravity until it reaches its terminal velocity, at which the force of air resistance equals the force of gravity, resulting in a constant speed.
To find the speed at the end of 3700 m, we first need to calculate the terminal velocity. We can use the equation for the drag force on a sphere:
Drag force = (1/2) * ρ * A * Cd * v²
where
ρ = density of air (1.2 kg/m³)
A = cross-sectional area of the raindrop (π * r²)
Cd = drag coefficient (0.45)
v = velocity of the raindrop
At terminal velocity, the drag force is equal to the weight of the raindrop:
Drag force = Weight of raindrop
(1/2) * ρ * A * Cd * v² = m * g
Substituting the values and rearranging the equation, we can solve for v:
v = √[(2 * m * g) / (ρ * A * Cd)]
Substituting the given values, we have:
v = √[(2 * (6.614 × 10⁻⁸ kg) * (9.8 m/s²)) / ((1.2 kg/m³) *
(π * (0.00175 m)²) * (0.45))]
Calculating the terminal velocity, we find:
v ≈ 8.136 m/s
Since the raindrop has already reached terminal velocity after falling about 30 m, its speed at the end of 3700 m will be the same as the terminal velocity:
Speed at the end of 3700 m ≈ 8.136 m/s
Therefore, the speed of the raindrop at the end of 3700 m when there is air drag is approximately 8.136 m/s.
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