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

For a refined model, we start with the given equation for acceleration:

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

Rearranging the equation, we separate the variables:

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

Next, we integrate both sides:

$\int \frac{1}{g - 0.1v} dv = \int dt.\n$

Doing the left side involves using a natural logarithm: $-\frac{10}{1} \ln |g - 0.1v| = t + C$

This gives: $|g - 0.1v| = e^{-\frac{t}{10} + C}$

The constant can be simplified to: $g - 0.1v = ke^{-\frac{t}{10}}$ To find k, we can apply an initial condition, for example, at $t = 0$, if we assume the initial velocity was zero: $g - 0.1(0) = k \Rightarrow k = g.$ Substituting back into the equation gives: $g - 0.1v = ge^{-\frac{t}{10}}$ Rearranging for v yields: $v = 10(g - ge^{-\frac{t}{10}}).$

As $t$ approaches large values:

1. For Andy's model:
   The velocity approaches $2g$ ms⁻¹, indicating that the ball accelerates under gravity without any influence of air resistance.

2. For Amy's refined model:
   As $t$ becomes large, the exponential term $e^{-\frac{t}{10}}$ approaches zero, leading the equation to simplify to: $v \approx 10g.$
   This suggests that while Amy's model initially takes into account a reduction in acceleration due to the influence of velocity, at large times, the increase in velocity stabilizes, but still results in a lower terminal velocity than the simple model. Thus, Amy's model reflects the presence of air resistance affecting the ball’s acceleration and terminal velocity.