# Motion of the skydiver

• imy786

## Homework Statement

A skydiver makes a controlled descent from an aeroplane to the ground. He
undergoes a number of seconds of freefall at his terminal speed before opening
his parachute, after which he decelerates to a new terminal speed. He continues to
fall at this speed for some time before landing on the ground.
Describe the motion of the skydiver starting from the moment he leaves the
aeroplane and finishing shortly before he hits the ground. You should divide the
descent into a number of phases. For each phase identify the vertical forces acting
upon the skydiver, and describe his motion using Newton’s laws. Where it is
useful to do so, as an approximation, you may treat the skydiver as behaving like
a sphere during his descent.
Remember to define the frame of reference that
you are using to describe the motion. You may ignore any horizontal forces and
motions.

F=MA
Newtons laws

## The Attempt at a Solution

minimum speed at start, increasing speed as time passes by. Until he hits the ground, where this is max speed. Vertical force going downwards increases as time goes by and upward vertical force decreases.

phase A= skydiver jumps out of plane traveling at speed till reaches terminal velocity
phase B = opens parachute , skydiver decelaretes to new terminal velocity
phase C= skydiver continues at this speed till reaches ground

phase A: person jumps out, accelerates until reaches termainal velocity, then a = o.

Person opens chute...
Right after the chute is opened, acceleration up

phase B: drag and air resistance are the same. The chute changes the profile of the falling object. The shape of the object changes, so the drag force will change as well.

During the entire fall, there are only two vertical forces, gravity and the drag force. The drag depends upon several aspects of the falling object.

phase A:
downward force- gravity,
upward force- air resistance

phase B:
downward force-gravity
upwward force= drag and air resistance

phase C:
downward force=gravity
upward force=no upward force when reached ground

?is this all ok?]

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imy786 said:
minimum speed at start, increasing speed as time passes by. Until he hits the ground, where this is max speed. Vertical force going downwards increases as time goes by and upward vertical force decreases.
For simplicity, let us just consider the vertical motion and speed. Are you sure that his speed continues increasing until he hits the ground? Are you also sure about your analysis of the forces involves? What does the term terminal velocity mean to you?

terminal velocity= maximum velocity, so therefore his speed reachese a limit while he is sky diving, at is at constant speed during some time before he reaches the ground.
This is to say acceleation will be constant when speed reaches max velocity.
So at that time vertical motion upward and downward will be equal.

??is this correct...any more details i can add

"This is to say acceleation will be constant when speed reaches max velocity."

Are you sure about this? If his velocity is not increasing, how can he still be accelerating?

after speed reaches maximum veloity speed will remain contstant therefore...no accelration.

imy786 said:
after speed reaches maximum veloity speed will remain contstant therefore...no accelration.

That's correct.

thanks Hage567...

well nearly there...just need to complete it.

Describe the motion of the skydiver starting from the moment he leaves the
aeroplane and finishing shortly before he hits the ground.

[i think this is kinda done]

You should divide the descent into a number of phases. For each phase identify the vertical forces acting upon the skydiver, and describe his motion using Newton’s laws. Where it is useful to do so, as an approximation, you may treat the skydiver as behaving like a sphere during his descent.

[still need to do]

Dividing the descent into a number of phases.

phase A= leaves the plane till reaches max velocity
phase B= terminal velocity till just hits the ground

any more phases that can be added?

would phase A be parachute opehing?

phase A = be parachute opehing
phase B= leaves the plane till reaches max velocity
phase C= terminal velocity till just hits the ground

is this corect?

The problem states that the person falls at a constant velocity for a few seconds before opening the chute. So it looks like you need a "phase" before your stated phase A.

phase A= skydiver jumps out of plane traveling at speed till reaches terminal velocity

phase B = opens parachute , skydiver decelaretes to new terminal velocity
phase C= skydiver continues at this speed till reaches ground

any more phases that could be added?

I think phase A can be broken down further.

For each phase identify the vertical forces acting upon the skydiver-

phase A:
downward force- gravity,
upward force- air resistance

phase B:
downward force-gravity
upwward force=thrust and air resistance

phase C:
downward force=gravity
upward force=no upward force when reached ground

are these vertical forces correct?

Phase B- thrust is an unusual term here, what do you mean.
Also, did you take this into account for your description of phase A/phase B: the person jumps out, reaches terminal velocity, falls at this rate for a few s, then opens the chute.

thrust its a guess...

What changes when the person opens the chute?

speed increases going upwards,
upward force increases more then downward force.

speed increases going upwards,
So what you are saying is that just after the chute opens, the person starts moving faster? does that make sense?

speed is a scalar

As the velocity vector becomes more downward, the drag force starts to act more and more in the upward direction, the skydiver continues to accelerate downward (but does not accelerate downward quite as quickly as -g). As the skydiver approaches the terminal velocity, the net force becomes smaller and smaller (and the divers motion becomes almost uniform).

As the velocity vector becomes more downward
I assume you mean that as the magnitude of the velocity vector increases...
Since, the drag is proportional to v or v^2, not sure how you are modeling this, the drag increases with v, yes.
Not sure where we are at now... Can you restate any questions you still have.

For each phase identify the vertical forces acting upon the skydiver, and describe his MOTION using Newton’s laws.

You may ignore any horizontal forces and motions

----------------------------------------------------------------------
phase A= skydiver jumps out of plane traveling at speed till reaches terminal velocity
phase B = opens parachute , skydiver decelaretes to new terminal velocity
phase C= skydiver continues at this speed till reaches ground

I think phase A can be broken down further...?

phase A:
downward force- gravity,
upward force- air resistance

phase B:
downward force-gravity
upwward force= drag and air resistance

phase C:
downward force=gravity
upward force=no upward force when reached ground

Now the question also states we may, can we still add horizonatal forces?

Okay, your phase A: person jumps out, accelerates until reaches v_t, then a = o. Person opens chute...
Right after the chute is opened, is the acceleration up or down?

phase B: drag and air resistance are the same. The chute changes the profile of the falling object. The shape of the object changes, so the drag force will change as well.

During the entire fall, there are only two vertical forces, gravity and the drag force. The drag depends upon several aspects of the falling object.

u explained saying until it reaches v_t = what does this mean?
do you mean
vector v (t)

sorry v_terminal

phase A= skydiver jumps out of plane traveling at speed till reaches terminal velocity
phase B = opens parachute , skydiver decelaretes to new terminal velocity
phase C= skydiver continues at this speed till reaches ground

phase A: person jumps out, accelerates until reaches termainal velocity, then a = o.

Person opens chute...
Right after the chute is opened, acceleration up

phase B: drag and air resistance are the same. The chute changes the profile of the falling object. The shape of the object changes, so the drag force will change as well.

During the entire fall, there are only two vertical forces, gravity and the drag force. The drag depends upon several aspects of the falling object.

phase A:
downward force- gravity,
upward force- air resistance

phase B:
downward force-gravity
upwward force= drag and air resistance

phase C:
downward force=gravity
upward force=no upward force when reached ground

?is this all ok?

I have to still be picky here. For instance phase A is composed of an accelerating portion of the motion and then a portion where a is zero. (the person reaches v_terminal)It is not clear to me that you see this as two distinct motions. Likewise for the other phases.
*If this is clear to you, can you state something about the net force on the system at each portion of the motion.*

phase A= skydiver jumps out of plane traveling at speed , accelrating at a=g
phase B= traveling at terminal velocity , a=0
phase C = opens parachute , skydiver decelaretes to new terminal velocity
phase D= skydiver continues at this speed till reaches ground

phase C: drag and air resistance are the same. The chute changes the profile of the falling object. The shape of the object changes, so the drag force will change as well.

During the entire fall, there are only two vertical forces, gravity and the drag force. The drag depends upon several aspects of the falling object.

phase A and B:
downward force- gravity,
upward force- air resistance

phase C:
downward force-gravity
upwward force= drag and air resistance

phase D:
downward force=gravity
upward force=no upward force when reached ground

OKay, now it is unclear to me as to whether we are to consider air drag on the person (chute closed). If so, which is quite reasonable, for phase A a will not equal g.

now it is unclear to me as to whether we are to consider air drag on the person (chute closed).

question only gives this info regarding what to be neglible-

You may ignore any horizontal forces and motions,

It seems reasonable to assume that air drag does act on the person (chute closed). Hence, a does not equal g during the first portion of motion.

phase A= skydiver jumps out of plane traveling at speed , acceleatating
phase B= traveling at terminal velocity , a=0
phase C = opens parachute , skydiver decelaretes to new terminal velocity
phase D= skydiver continues at this speed till reaches ground

phase C: drag and air resistance are the same. The chute changes the profile of the falling object. The shape of the object changes, so the drag force will change as well.

During the entire fall, there are only two vertical forces, gravity and the drag force. The drag depends upon several aspects of the falling object.

phase A and B:
downward force- gravity,
upward force- air resistance

phase C:
downward force-gravity
upwward force= drag and air resistance

phase D:
downward force=gravity
upward force=no upward force when reached ground

when does a=g then...when it reaches ?
well gravitoanl force = acceleration so when the sky diver is accerating he must be accelrating at g...what other quantity??

is this correct Rob??

The only time that the person accelerates at g is when air drag is zero. Where does this occur?
All phases look better except D. When the person reaches the ground, the motion is over;lets ignore that part. With the chute open and falling at v = v_terminal, what forces act?

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