The Physics of Landing Softly
By: Sam Yu
Why does a parachute slow decent rate?
When we begin considering the physics behind parachutes, we must start with some definitions.
Drag Force: The upward force felt by a falling object due to several factors given by the equation, Force Drag = 1/2(Density)(Area)(Drag Coefficient)(Velocity)^2
Density: Density of the air, usually 1.27kg/m^3 at STP
Area: The cross sectional area of the falling object, perpendicular to ground
Drag Coefficient: A unit less quantity that is dependent on the type of material used in the parachute
Velocity: The rate of decent of the falling object at a certain moment
Terminal Velocity: The instance of when the upward drag force equals the downward gravitational force. The net force of the system is zero.
These factors all contribute to the upward force experienced by a falling object, while the only downward force experienced by the object is gravity. With or without a parachute, a falling object will continue to accelerate until it reaches terminal velocity.
Looking at the equation, Force Drag = 1/2(Density)(Area)(Drag Coefficient)(Velocity)^2 we can see that parachutes will give us a much bigger area and a greater Drag Coefficient than say a cube or a person. Effectively lowering the needed velocity needed to reach a drag force great enough to create a net force of zero on a falling object.
I decided to test the terminal velocity and impact force experienced by a falling toy when attached to a parachutes of different sizes. The variables that I was testing are as follows:
Coefficient of Drag
Velocity at time of impact
Total time of impact
Force of Impact
My constant was the mass of my toy
Several assumptions were made in my model.
1. The time of impact was the same for all collisions
2. The impact of the toy with the ground was perfectly inelastic
3. The toy had negligible drag
I ran three experiments determine the values of my variables, the first experiment was to determine the time it took for impact to occur and confirm that my toy had negligible drag force for the three meters it took for it to fall.
By using a force plate running 500 samples a second, I was able to find the average time it took to for newton's third law to occur. The time for a elastic collision to occur was on average:
Avg Impact Time: .015 seconds
Free Fall Acceleration Graphs
Using video analysis with logger pro, I was able to determine that the object had negligible drag on its own and that the acceleration was unchanged falling from a height of 3 meters.
Avg calculated gravitational constant from 3 meters: 9.957m/s^2
Avg velocity at moment of impact: 7.65m/s
To calculate the moment of impact, a slightly altered version of Force = M * A can be used. Force = (Mass) * (Delta Velocity)/( Delta Time) where our discovered variables are (.153kg), (.015s), (7.65m/s):
Now that we know that the impact time, the toy has negligible drag and the parachute-less impact force. We can solve for situations with parachutes and compare the impact forces.
Again, remember our equation:
Force Drag = 1/2(Density)(Area)(Drag Coefficient)(Velocity)^2
In this experiment our parachute area is 16in x 16in(which needs to be converted into meters). Since we are looking for the terminal velocity and its impact, our net force when the toy is falling needs to be zero. Therefore we can set our force of gravity equal to our drag force to get the equation:
(mass)(gravity) = 1/2(Density)(Area)(Drag Coefficient)(Velocity)^2
If we take our mass, .153kg and multiply by the gravitational constant 9.81m/s^2 we can get our downward and drag forces.
Downward and Drag: 1.5Newtons
Next we need to find our drag Coefficient for our parachute and terminal velocity. I found terminal velocity using video analysis and from there calculated the drag coefficient.
Taking a look at the graph, we can see that our toy reached a terminal velocity of -3.493m/s at about .5s into the fall.
Now that we have all our variables we can solve for the drag coefficient of our parachute.
Lastly we need to solve for our force of impact, again we use the same equation with the variables we solved for previously(.153kg),(-3.493m/s),(.015s):
Force = (Mass) * (Delta Velocity)/( Delta Time)
We can see that the force of impact with a 16in x 16in parachute is less than half of the impact force of the object in free fall(78.03) Newtons
Experiment 3 was performed very similarly to the second. The only change made was the size of the parachute, 24in x 24in. Since the same material was used there is no need to recalculate for the drag coefficient. Now, if we go to the video analysis we can see that again our terminal velocity has decreased to -2.422m/s. We can see that increasing our chute by a 1/3 decreased our velocity by 1/3.
Finally, we calculate our last impact force using the equation with our found variables, (.153kg),(-2.422m/s),(.015s):
Force = (Mass) * (Delta Velocity)/( Delta Time)
Remember that we were looking for several variables and how they influence each other:
Ex:1 Ex:2 Ex:3
Size of Parachute: N/A 16in x 16in 24in x 24in
Area of Parachute: N/A .1651m^2 .3716m^2
Coefficient of Drag: N/A 1.174 1.174
Velocity at time of impact: -7.65m/s -3.493m/s -2.422m/s
Total time of impact: .015s .015s .015s
Drag Force Terminal: N/A 1.5N 1.5N
Force of Impact: 78.03N 35.63N 24.7N
Height: 3 meters
I found that when considering parachute size. Increasing the size by 1/3 decreased the terminal velocity by 1/3 and the impact force by 1/3.