Estimated terminal velocity significantly less than estimated impact velocity

[FEllen]

Homework Statement

I am currently conducting an EEI in which I am looking at the effects of increased mass, surface area and air pressure on terminal velocity. In my experiment, I dropped a parachute of 0.36m^2 from a height of 3m with a mass of 0.01kg. My result was that it took 3.036 seconds to fall, giving an average velocity of 0.988 m/s. Assuming the object did not reach terminal velocity, and exprienced constant acceleration, the impact velocity should then be around 2m/s (double the average). However, the estimated terminal velocity of the object I calculated was 0.79m/s. This is even less than the average velocity - which should theoretically be impossible. This has happened with most my results! Anyone know what I could be doing wrong? Even taking error into account does not decrease impact velocity enough.

Homework Equations

Drag formula:
F= 1/2*ρCAv^2
Where F is the drag force;
ρ is the air density;
C is the drag coefficient;
A is the area of the parachute;
And v is the velocity through the air

Thus making terminal velocity:
v=√(2W/ρCA)

The Attempt at a Solution

Air density was estimated to be 1.171kg/m^3, and the drag coefficient of the parachute was 0.75.
v=√(2*9.81*0.01/(1.171*0.75*0.36))
v= 0.79m/s

 

[cepheid]

Welcome to PF!,

What is an EEI?

How do you know that the drag coefficient of the parachute is 0.75?

 

[PeterO]

Homework Statement

I am currently conducting an EEI in which I am looking at the effects of increased mass, surface area and air pressure on terminal velocity. In my experiment, I dropped a parachute of 0.36m^2 from a height of 3m with a mass of 0.01kg. My result was that it took 3.036 seconds to fall, giving an average velocity of 0.988 m/s. Assuming the object did not reach terminal velocity, and exprienced constant acceleration, the impact velocity should then be around 2m/s (double the average). However, the estimated terminal velocity of the object I calculated was 0.79m/s. This is even less than the average velocity - which should theoretically be impossible. This has happened with most my results! Anyone know what I could be doing wrong? Even taking error into account does not decrease impact velocity enough.

Homework Equations

Drag formula:
F= 1/2*ρCAv^2
Where F is the drag force;
ρ is the air density;
C is the drag coefficient;
A is the area of the parachute;
And v is the velocity through the air

Thus making terminal velocity:
v=√(2W/ρCA)

The Attempt at a Solution

Air density was estimated to be 1.171kg/m^3, and the drag coefficient of the parachute was 0.75.
v=√(2*9.81*0.01/(1.171*0.75*0.36))
v= 0.79m/s

Could it be that the parachute takes a little while to fully deploy properly?
If so the system would be simply falling, at first, with an acceleration of 9.8, but that acceleration would quickly reduce under the air resistance. Even so the falling system may reach a higher speed before the fully opened, and stable, parachute resulted in the final , terminal velocity.

A video of the experiment could be useful. You could analyse the video frame by frame to see what is going on. Be careful if you use the "high speed shutter" facility to get nice clear images. Early cameras with that feature did not "take" regularly spaced [time] images. It seemed that when operating at 25 frames per second, but with the 0.001 sec shutter, the image was taken at almost random times during each 0.02 second period.

Who knows, the modern i-phone may do an excellent job?

 

[FEllen]

I got the estimated coefficient of drag from http://www.aeroconsystems.com/chutes/drag_calculator.htm (as it was a parasheet made out of garbage bag). An EEI is an Extended Experimental Investigation - really just a long assignment. It is possible that it took some time to deploy, but I don't think that could account for all the difference.

 

[PeterO]

I got the estimated coefficient of drag from http://www.aeroconsystems.com/chutes/drag_calculator.htm (as it was a parasheet made out of garbage bag). An EEI is an Extended Experimental Investigation - really just a long assignment. It is possible that it took some time to deploy, but I don't think that could account for all the difference.

The whole purpose of the parachute is to prevent the falling object from having a constant acceleration, so you estimate of the final velocity as twice the average has no basis.

Neither, I suspect, is any theoretical terminal velocity.

I think you need to devise a method of measuring the velocity over the last metre, and checking whether it is constant [ie. the terminal velocity].

 

[FEllen]

Okay, so terminal velocity increases as the object velocity increases, meaning it is not a constant deceleration (as the force goes up). Any ideas on how to calculate the impact velocity otherwise? Also, that does not answer why the average velocity would be greater than the terminal velocity...

 

[PeterO]

Okay, so terminal velocity increases as the object velocity increases, meaning it is not a constant deceleration (as the force goes up). Any ideas on how to calculate the impact velocity otherwise? Also, that does not answer why the average velocity would be greater than the terminal velocity...

I think you had better address the question of what Terminal Velocity is - judging by what I highlited RED above.

Part of the design of your EEI is to measure the terminal velocity.

If an object increased its speed to 3 m/s while the parachute began to work, then slowed to 1.5 m/s before landing, it could well have a a higher average speed than its terminal velocity.