A skydiver [m=95kg] jumps out of an airplane and opens her parachute 45 seconds later. After another 15 seconds, the skydiver begins to descend gently at a constant speed of 7.8m/s. Draw a freebody diagram of the skydiver just after she's jumped out the airplane AND 1 minute after she has jumped out. Calculate the acceleration and the force of friction on the skydiver after she's fallen for 1 minute.

freebody: after she jumps: gravity down.

in one min: gravity down, air resistance up

if she is at constant speed, acceleration is zero, and her friction is equal to her weight.

To draw the free body diagrams of the skydiver at two different instances, we need to understand the forces acting on her.

1. Just after jumping out of the airplane:
Here is a breakdown of the forces acting on the skydiver just after she has jumped out of the airplane:
- Weight (mg): This is the force due to gravity acting vertically downwards. Its magnitude can be calculated using mg, where m is the mass of the skydiver (95 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Air resistance/ drag force (Fd): This force opposes the motion of the skydiver and acts vertically upwards. Its magnitude depends on various factors such as the shape of the skydiver's body and the speed of descent. However, since we do not have specific information about these factors, we can represent it with an unknown force Fd.

2. One minute after jumping out of the airplane:
At this point, the skydiver has opened her parachute and is descending gently at a constant speed of 7.8 m/s. In this case, the forces acting on her are:
- Weight (mg): As before, this force acts vertically downwards.
- Air resistance/ drag force (Fd): This force opposes the motion of the skydiver but is now equal in magnitude to the weight (mg). This occurs when the skydiver reaches terminal velocity, which is the maximum speed she can achieve due to air resistance balancing her weight.

Now let's calculate the acceleration and the force of friction after one minute of falling:

Acceleration (a) after falling for 1 minute:
We know that the skydiver descends at a constant speed of 7.8 m/s. When an object moves at a constant speed, its acceleration is zero. Therefore, the acceleration of the skydiver after one minute of falling is 0 m/s^2.

Force of friction (Ff) after falling for 1 minute:
Since the acceleration is zero, we can conclude that the net force acting on the skydiver is also zero. Therefore, the force of air resistance (Fd) must be equal in magnitude but opposite in direction to the weight (mg).

Since we know that Fd is equal to mg, and m = 95 kg, and g = 9.8 m/s^2, we can calculate the force of friction (Ff):
Ff = Fd = mg = 95 kg * 9.8 m/s^2 = 931 N.

Therefore, the force of friction on the skydiver after she has fallen for 1 minute is 931 Newtons.