Question

A skydiver of mass 65 kg has jumped out of a plane two miles above he surface of the earth. After 50 s, he has reached terminal velocity, meaning he is no longer accelerating. What is the force of the air on the skydiver's body? g = m/s^2. *​

Answer 1

The force of the air on the skydiver's body is:

`F_(air)=637.7\: N`

Step-by-step explanation:

Let's recall that when an object reached terminal velocity, the force of the air acting is equal to the weight of the object, so the net force will be zero, that is why the object falls at a constant speed.

Then we have:

`F_(air)=W=mg`

where:

m is the mass of the skydiver (65 kg)

g is the gravity acceleration (9.81 m/s²)

Therefore, the force of the air on the skydiver's body is:

`F_(air)=65*9.81=637.7\: N`

I hope it helps you!