Question

You drop the same ball of mass 5 kilograms out of the same airplane window at the same height, except this time you assume a drag force proportional to the ball's velocity, using a proportionality constant of 3 and the ball reaches terminal velocity. Solve for the distance fallen as a function of time. How long does it take the ball to reach the ground?

Short answer

Calculate velocity and solve differential equations for distance fallen. Use numerical methods to find "time to reach the ground" based on distance.

Step-by-step solution

Understanding Terminal Velocity

When an object falls under gravity with air resistance, it reaches a terminal velocity where the drag force equals the gravitational force. The forces balance as follows: \[ F_{ ext{drag}} = F_{ ext{gravity}} \] The gravitational force is \( F_{\text{gravity}} = mg \), where \( m = 5 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). The drag force is given by \( F_{\text{drag}} = -kv \), where \( k = 3 \text{ s^{-1}} \) and \( v \) is the velocity.

Calculating Terminal Velocity

Set the magnitude of the drag force equal to the gravitational force at terminal velocity:\[ mg = kv_t \] where \( v_t \) is the terminal velocity. Plug in the known values: \[ 5 \times 9.8 = 3v_t \] Solving for \( v_t \) gives:\[ v_t = \frac{49}{3} \approx 16.33 \, \text{m/s} \]

Setting Up the Differential Equation

The net force on the ball is the sum of the gravitational force and the drag force, leading to the differential equation:\[ m\frac{dv}{dt} = mg - kv \]Simplifying, we find:\[ \frac{dv}{dt} = g - \frac{k}{m}v \]This is a separable first-order differential equation.

Solving the Differential Equation for Velocity

To solve \( \frac{dv}{dt} = g - \frac{k}{m}v \), separate variables:\[ \frac{dv}{g - \frac{k}{m}v} = dt \]Integrate both sides:\[ \int \frac{1}{g - \frac{k}{m}v} \, dv = \int dt \]The left side results in a natural logarithm:\[ -\frac{m}{k} \ln |g - \frac{k}{m}v| = t + C \]Solve for \( v \):\[ v(t) = \frac{mg}{k} (1 - e^{-\frac{k}{m}t}) \]

Integrating to Find Distance Fallen

To find the distance fallen, integrate the velocity function:\[ x(t) = \int v(t) \, dt = \int \frac{mg}{k}(1 - e^{-\frac{k}{m}t}) \, dt \]Split the integral:\[ x(t) = \frac{mg}{k}t + \frac{m^2g}{k^2} e^{-\frac{k}{m}t} + C' \] Apply initial condition \( x(0) = 0 \) to solve for \( C' \). Set \( C' = -\frac{m^2g}{k^2} \) to satisfy initial conditions.

Determine Time to Reach Ground

Assuming the ball falls from a height \( h \), solve \( x(t) = h \) to find time. Substitute: \[ \frac{mg}{k}(t + \frac{m}{k}e^{-\frac{k}{m}t}) = h \] Use trial, error, or numerical methods to solve for \( t \). For example, let's assume \( h = 1000 \text{ m} \), substitute and solve approximately, using estimates or numerical tools.

Key concepts

Differential Equations

When analyzing motion under the influence of forces, differential equations play a crucial role. A differential equation is an equation that involves a function and its derivatives. In the context of terminal velocity, the motion of the falling ball is described by a differential equation because the velocity is changing over time due to forces acting on the object.

The specific differential equation we encounter here is derived from Newton's Second Law, which states that the sum of forces is equal to mass times acceleration. For the ball experiencing both gravity and drag force, the equation becomes:

• Mass (\( m \)) times the change in velocity over time (\( \frac{dv}{dt} \)) equals gravitational force minus drag force.
• The resulting differential equation: \[ m\frac{dv}{dt} = mg - kv \]

Separable differential equations like this one can be solved by integrating both sides, allowing us to find velocity as a time function.

Gravitational Force

Gravitational force is one of the primary forces acting on the ball as it falls. This force pulls the ball towards the Earth, accelerating it downwards by the value of acceleration due to Earth's gravity, which is approximately \( 9.8 \, \text{m/s}^2 \).

For an object with mass \( m \), gravitational force (\( F_{gravity} \)) is calculated using the formula:

• \( F_{gravity} = mg \)
• In this context, \( m = 5 \) kilograms, leading to a force of \( 49 \) newtons.

This constant force acts downward, increasing the ball’s velocity until overcoming the drag force.

Drag Force

As the ball falls, it encounters air resistance, known as drag force. This force opposes the motion of the ball and increases with velocity.

The drag force in this context is proportional to the velocity (\( v \)) of the ball. The relationship is given by:

• \( F_{drag} = -kv \)
• \( k \) is the proportionality constant, in this case \( 3 \) s⁻¹, indicating how strongly the drag force resists the ball’s velocity.

As the ball speeds up, the drag force increases until it balances the gravitational force, leading to terminal velocity.

Velocity as a Function of Time

The velocity of the falling ball is not constant; it changes over time due to the forces acting on it. By solving the relevant differential equation, we can derive a formula to describe this velocity.

The equation derived is:

• \( v(t) = \frac{mg}{k} (1 - e^{-\frac{k}{m}t}) \)
• This formula accounts for the increasing velocity due to gravity and the opposing resistance from the drag force.
• As time (\( t \)) increases, the exponent \( e^{-\frac{k}{m}t} \) approaches zero, indicating that the velocity approaches a constant value known as terminal velocity.

The terminal velocity occurs when the gravitational force and drag force are equal, resulting in no net force and thus no further acceleration.

Distance Fallen Calculation

To find out how far the ball has fallen at a given time, we need to integrate the velocity function over time. This integration provides a formula for the distance fallen as a function of time.

The process results in:

• \( x(t) = \left( \frac{mg}{k} \right) t + \frac{m^2g}{k^2} e^{-\frac{k}{m}t} + C' \)
• The integration constant \( C' \) can be found using the initial condition that at \( t=0 \), the distance \( x(0) = 0 \).
• Solving for \( C' \) helps ensure the formula accurately models the distance from when the ball first began to fall.

Solving this equation for a specific distance allows us to determine how long it takes for the ball to reach that distance, such as the ground in this scenario.