Question

A 50 kg parachutist jumps out of an airplane at a height of 1000m. The parachute opens, and the jumper lands on the ground with a speed of 5 m/s. What is the average force of air resistance during the descent?

Answer 1

The force is equal to 489.87 [N]

Step-by-step explanation:

We need to make a free body diagram whe we can see the forces acting over the parachutist. In the attached image we can see the sketch of the free body diagram.

First we use the analysis of forces equal to mass by acceleration to be able to determine the equation with the term of the resistance force.

Then we have to find the acceleration to be able to apply it in the equation of the resistance force of the parachute. The acceleration can be found using the following kinematics equation

`\\v_(f)^(2)=v_(0)^(2)+2*a*y\\ where\\v_(f) = final velocity = 5 [m/s]\\v_(o)= initial velocity = 0 [m/s]\\y=elevation =1000[m]\\`

Note: here the acceleration is positive because the direction of the acceleration is towards the movement

Therefore the resistance force is 489.87 [N] or 49.93[kg-f]

Note: We can see that the resistance force of the parachute is almost equal to the equivalent of the weight of the parachutist, although its value is lower in this way it is explained by the fact that the system is not in balance and the person descends.