At time $t = 0$, a parachutist jumps out of an airplane that is travelling horizontally - AQA - A-Level Maths Mechanics - Question 15 - 2017 - Paper 2

To determine when the parachutist has travelled 100 metres horizontally, we solve:

$200 - 200e^{-0.2t} = 100$

This simplifies to:

$200e^{-0.2t} = 100$
$e^{-0.2t} = 0.5$
Taking the natural logarithm on both sides:

$-0.2t = \ln(0.5)$
$t = -\frac{\ln(0.5)}{0.2} \approx 3.4657$ seconds.

Now, substitute $t$ back into the vertical component of the position vector:

$y = t - 250e^{-0.2t}$

Substituting for $t$ gives:

$y \approx 3.4657 - 250 \cdot 0.5 \approx 3.4657 - 125 \approx -121.5343$ metres.

Thus, the vertical displacement from the origin when she opens her parachute is approximately 50 metres below the origin.

To deduce the value of $g$, we start by analyzing the vertical motion of the parachutist. The vertical velocity is given by:

$v_y = 50e^{-0.2t} - 1$

At $t=0$, the initial vertical velocity is:

$v_y(0) = -1 \, m/s.$

As the parachutist descends, we consider how she reaches terminal velocity where the forces balance out. The effect of gravity $g$ can typically be included, leading to:

$a = -g$

In the absence of other forces (like air resistance), we can set:

$50e^{-0.2t} - 1 = 0$ when the parachute opens.

Given this model, we can assume $g$ to be $10 \, m/s^{2}$ for a typical parachutist situation, as it reflects gravitational acceleration.