Using parachutes A relief organisation drops food parcels by parachute from a helicopter - Edexcel - GCSE Physics - Question 5 - 2017 - Paper 1

To calculate the velocity, we use the equation of motion:

where:

• $v$ is the final velocity,
• $u$ is the initial velocity (0 m/s since it starts from rest),
• $g$ is the acceleration due to gravity (10 m/s²), and
• $t$ is the time of fall (1.2 s).

Plugging in the values:

The velocity of the food parcel after falling for 1.2 s is 12 m/s.

In the diagram, the forces acting on the food parcel are:

• Weight of the parcel = 80 N downwards
• Air resistance acting upwards = 105 N

To find the resultant force, we subtract the upward force from the downward force:

Thus, we select B for the resultant force on the food parcel.

The graph shows a positive correlation between the velocity of the parachute and the air resistance acting on it. As the velocity increases, the air resistance also increases. This indicates that the air resistance is dependent on the speed of the parachute.

Initially, when the food parcel is dropped, only the gravitational force acts on it, causing it to accelerate downwards. As it falls, the speed increases, leading to an increase in air resistance. Eventually, the upward air resistance force may equal the downward weight of the parcel, resulting in a constant velocity (terminal velocity). In this scenario, the forces are balanced, meaning no further acceleration occurs.