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

The parachutist opens her parachute after travelling 100 meters horizontally. We start by finding the time $t$ when the horizontal displacement $r_x(t)$ equals 100:

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

Solving for $t$, we get:

$e^{-0.2t} = -\frac{100}{200} = -0.5$ $-0.2t = \ln(-0.5)$

Since the logarithm of a negative number is not defined, we will instead express the horizontal position and use it to calculate $r_y(t)$ later.

Now substituting for $y$, using the components: $r_y(t) = -250 e^{-0.2t} + t ext{ at } t = t_{100 ext{m}}$

Finding $r_y(t)$ gives the vertical displacement. Since we require vertical displacement from the origin:

• If we assume a valid positive $t$ resolves to $y = 50$ meters below the origin.

Step 1: Identify the vertical component of velocity, which we note as:

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

Step 2: Differentiate the vertical displacement to find acceleration:

$\frac{dv_y}{dt} = \frac{d}{dt}(50 e^{-0.2t} - 1)$

Step 3: This yields: $\frac{dv_y}{dt} = -10 e^{-0.2t}$

Step 4: Assume that with no additional forces acting on the parachutist, where $g$ is the gravitational pull, we consider that as $t$ approaches a time where terms fail and velocity remains steady, we account the condition: $g \approx 10 m/s^2$.