A ball is released from a great height so that it falls vertically downwards towards the surface of the Earth - AQA - A-Level Maths Pure - Question 17 - 2021 - Paper 2

Using the refined model given by Amy, we start with the equation for acceleration:

$a = \frac{dv}{dt} = g - 0.1v$

Rearranging gives: $\frac{dv}{dt} + 0.1v = g$

This is a first-order linear ordinary differential equation. We can solve it by finding an integrating factor:

Let: $P(t) = e^{\int 0.1 \, dt} = e^{0.1t}$

Multiplying through by the integrating factor: $e^{0.1t} \frac{dv}{dt} + 0.1e^{0.1t} v = ge^{0.1t}$

Integrating both sides: $\int \left( e^{0.1t} v \right)' dt = \int g e^{0.1t} dt$

The left side simplifies to: $e^{0.1t} v = \frac{g}{0.1} e^{0.1t} + C$ Where $C$ is the constant of integration. Dividing both sides by $e^{0.1t}$ gives: $v = \frac{g}{0.1} + Ce^{-0.1t}$

Assuming at $t = 0$, $v = 0$, we find: $0 = \frac{g}{0.1} + C \Rightarrow C = -\frac{g}{0.1}$

Thus, we can substitute $C$ back into our equation: $v(t) = \frac{g}{0.1} (1 - e^{-0.1t})$

As $t$ becomes large, we consider the behavior of the velocity predictions from both models:

1. Simple Model (Andy’s prediction):

   • From the simple model, we derived that: $v = 2g \approx 19.62 \, \text{ms}^{-1}$
   • This indicates that the velocity approaches a constant value (2g) as the ball continues to fall without considering air resistance.

2. Refined Model (Amy’s prediction):

   • In Amy's model, as $t \to \infty$, the term $e^{-0.1t} \to 0$, leading to: $v = \frac{g}{0.1} \approx 98.1 \, \text{ms}^{-1}$
   • This shows that the velocity stabilizes at a higher value, accounting for the effects of air resistance.

In conclusion, Andy's model results in a fixed velocity value of 2g, while Amy's refined model accounts for increasing velocity influenced by air resistance, resulting in an asymptotic behavior as $t$ increases.