Dropped off of an airplane, into a wall.

[DavidC]

Ok, the next question is a little funnier. I was curious what scenario would create more impact force.

A) An airplane drops an object from 500 feet, which impacts into the ground (assume just solid, level granite).

B) An airplane drops an object from 500 feet, which falls 200 feet before hitting a wall (solid granite). The object then falls the remaining 300 feet to the ground and hits the ground.

I'm just curious if the total force hitting this object would be the same or different in A and B.

[Fels krone]

The 500 foot drop would be more. The kinetic energy of the object is defined by velocity squared, so the higher velocity would be much more impact force. It could be limited by its terminal velocity, but if that velocity was more than 150 mph it wouldnt have time to reach it.

[Sabrazack]

I believe acceleration slows as falling speed increases. This might have an effect here? But then again, maybe not on such a short fall.

[Silent A]

You have to fall about a kilometer or so to reach terminal velocity so 500 feet is not enough.

While it's true that the energy is proportional to the velocity squared, the impact force is proportional to the square root of the kinetic energy (assuming the ground acts like a spring), so the impact force is proportional to the speed.

Unfortunately, at these heights, air resistence is enough to significantly slow down the object but not enough to reach terminal velocity so it's impossible to calculate the impact velocity accurately without doing some calculus (which I don't have time for).

But if you ignore air resistence you'll find that in (b) the two impact forces together will be greater than the single impact force in (a), and if you think about the physics a little you'll realize that air resistence will make this difference even bigger because it will have a bigger effect for a 500 ft drop than on a 200 or 300 ft drop.

So more total force in case (b).

However, adding forces in this way doesn't have any physical meaning. You can only add forces when they occur at the same time.

[CORed]

[ QUOTE ]
I believe acceleration slows as falling speed increases. This might have an effect here? But then again, maybe not on such a short fall.

[/ QUOTE ]

In a vacuum, the acceleration of a falling object is constant. In atmosphere, wind resistance will eventually reduce the acceleration to zero. When this happens, the falling object is at terminal velocity. Wind resistance will also slow the forward speed of the object.

[Siegmund]

Depends on the object and the airplane -- but probably B, unless it's an extremely dense and streamlined object and an extremely slow-moving airplane.

Terminal velocity of a person, for instance, is 120mph. He will be slowing down from the instant he exits a typical airplane onward. (Accelerating downward, but losing more forward speed than he is gaining downward.)

EDIT TO ADD:

Looking at it another way -- if the ground is perfectly horizontal and the wall is perfectly vertical... the object strikes the ground at a similar speed after accelerating downward for 500 feet in both cases, *in addition to which* object B slams sideways into something.

[Matt R.]

500 ft = 152.4m

From a height of 152.4m, with initial velocity of zero, the object has final velocity
vf = sqrt(2g*152.4) = 54.7 m/s
F = dp/dt = m dv/dt
Assuming the object decelerates at a constant rate upon impact,
F = (54.7 m/s)*m/dt
where m is mass of object, and dt is amount of time it takes to stop the object (pretend that says delta t).

From 300 ft = 91.44 m
vf = sqrt(2g*91.44) = 42.3 m/s

And from 200 ft = 60.96 m
vf = 34.6 m/s

F(total) = 42.3*m/dt + 34.6*m/dt = 76.9*m/dt

76.9 > 54.7, so the total force impacting the object that "stops" once at 200 ft is greater than the total force impacting the object falling the full 500 ft without stopping. This is assuming the amount of time to decelerate the object is the same each time (which wouldn't quite be true, but I think this is a good estimate).

[Silent A]

[ QUOTE ]
F(total) = 42.3*m/dt + 34.6*m/dt = 76.9*m/dt

76.9 > 54.7, so the total force impacting the object that "stops" once at 200 ft is greater than the total force impacting the object falling the full 500 ft without stopping. This is assuming the amount of time to decelerate the object is the same each time (which wouldn't quite be true, but I think this is a good estimate).

[/ QUOTE ]

dt for 500ft >= dt for 300 ft >= dt for 200 ft

If you think about it, this would only make the difference even greater.

But again, adding the forces in case (b) has no physical meaning.

[Matt R.]

Yep, I initially thought the sum of the forces in (b) would be equal to the force from the "full drop", and after seeing they weren't equal I thought the differences in dt would account for this. But then I realized it only makes the difference greater.

Also, yeah, I agree summing the forces in (b) doesn't tell you much about the physics except for the total force acting on the object over the two impacts. Just answering the OP's hypothetical.

[Silent A]

[ QUOTE ]
Also, yeah, I agree summing the forces in (b) doesn't tell you much about the physics except for the total force acting on the object over the two impacts. Just answering the OP's hypothetical.

[/ QUOTE ]

I expected as much. I only mentioned it for emphasis and for the benefit of others.