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

Using the refined model, we start with the given equation for acceleration: $a = g - 0.1v$

To find the velocity $v$ at time $t$, we use: rac{dv}{dt} = g - 0.1v

This is a first-order linear differential equation. We can separate variables and write it as: rac{dv}{g - 0.1v} = dt.

Integrating both sides: rac{1}{0.1} ext{ln}|g - 0.1v| = t + C

Where $C$ is the constant of integration. Solving for $v$, we obtain: $|g - 0.1v| = 10e^{(t+C)}.$

By evaluating the constants, we find that: $v = 10g(1 - e^{-0.1t}).$

As $t$ approaches large values, we observe different behaviors in the two models:

1. In the simple model (Andy’s prediction), the velocity approaches $2g$ ms⁻¹ consistently without any upper limit since it is based purely on constant acceleration due to gravity.

2. In the refined model (Amy’s prediction), the velocity approaches a limit of $10g$ as $t o ext{large}$, indicating that as the resistance effect increases, the acceleration diminishes until the ball reaches a terminal velocity.

This shows a significant difference in predictions based on the modeling approach.