How Long Does It Take a Dropped Bag to Fall from a Helicopter?

[eddddddd]

1. The problem statement, all variables andi given/known data
A helicopter ascending at a velocity of 5.0m/s drops a bag at 105 m. How long does it take the bag to fall?

Homework Equations

X=vit+.5at^2

The Attempt at a Solution

I've tried the equation above but i can't solve for time

 

[haruspex]

1. The problem statement, all variables andi given/known data
A helicopter ascending at a velocity of 5.0m/s drops a bag at 105 m. How long does it take the bag to fall?

Homework Equations

X=vit+.5at^2

The Attempt at a Solution

I've tried the equation above but i can't solve for time

Show some more of your working. What did that equation give you when you plugged in the numbers?

 

[Jenab2]

Gravitational parameter of Earth,
GM = 3.986004418e+14 m³/s²

Radius of Earth,
R = 6371000 meters

Altitude of helicopter at bag release,
h₁ = 105 meters

Velocity of bag at release,
v₁ = +5 m/s

geocentric radius of bag at release,
r₁ = R + h₁
r₁ = 6371105 meters

The Vis Viva equation,
v₁ = √[GM(2/r₁ − 1/a)]

In a plunge orbit, the eccentricity is one, and therefore the apoapsis distance, d, is twice the semimajor axis, a.

a = d/2

And the Vis Viva equation can be rewritten as

d = [1/r₁ − v₁²/(2GM)]⁻¹

In the example problem, the apoapsis distance is found to be

d = 6371106.2729222 meters

The general solution for the time to fall in a plunge orbit is

t−t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

The time for the bag to rise from release (at t=t₋₁) to its apoapsis (at t=t₀) altitude is equal to the time for the bag to fall from the apoapsis back to the release altitude (at t=t₁). Thus,

t₁−t₀ = √[d/(2GM)] { √(r₁d−r₁²) + d arctan √(d/r₁−1) }
t₁−t₀ = 0.509168900 sec

t₁−t₋₁ = 2(t₁−t₀)
t₁−t₋₁ = 1.018337801 sec

The time for the bag to fall from the apoapsis altitude to the ground (r=R, t=t₂) is

t₂−t₀ = √[d/(2GM)] { √(Rd−R²) + d arctan √(d/R−1) }
t₂−t₀ = 4.652335898 sec

So the total time from bag release to bag hitting ground is
T = (t₂−t₀) + (t₁−t₀) = 5.161504798 sec

A derivation of the formula for the general solution for the time to fall in a plunge orbit can be found at

https://www.physicsforums.com/threads/the-time-to-fall-in-a-plunge-orbit.873612/post-6264578

 

[haruspex]

The Vis Viva equation,
v₁ = √[GM(2/r₁ − 1/a)]

That is irrelevant to the question in post #1, which is at a very modest altitude. We can take gravitational acceleration as the standard constant g.

 

[Jenab2]

Look closely. The altitude used in my treatment is 105 meters, which is what the question specified. The difference between my method and the approximate method is that I've provided the classically exact solution. No approximation necessary. Naturally, the approximation in this case is very close to the truth. But the asker might find it convenient to have the exact solution for other cases, in which the length of the fall is a substantial fraction of (or is greater than) the radius of Earth.

 

[haruspex]

Look closely. The altitude used in my treatment is 105 meters, which is what the question specified. The difference between my method and the approximate method is that I've provided the classically exact solution. No approximation necessary. Naturally, the approximation in this case is very close to the truth. But the asker might find it convenient to have the exact solution for other cases, in which the length of the fall is a substantial fraction of (or is greater than) the radius of Earth.

Sure, but your method is way beyond what is needed, and likely well beyond the level of the student. That makes it inappropriate.

And if it had been appropriate I wouid be asking you or a moderator to edit your post. We do not provide complete solutions, at least not straightaway. Please start with just providing hints and pointing out errors. See the Forum Rules.

 

[gmax137]

the classically exact solution. No approximation necessary

Well, you have neglected friction with the air, the rotation of the earth, the gravitational attraction from the moon, the non-spherical shape of the earth, and so on. So you have not provided an exact solution.

the approximation in this case is very close to the truth

please do not confuse

5.161504798 sec

with the "truth." All of the approximations you made would affect this result (leaving aside the question of significant digits: your answer assumes initial velocity is actually 5.000000000 m/sec and altitude is actually 105.0000000 meters).

If the question had been, "estimate the effect of assuming constant acceleration in determining the time to fall..." then your approach might be wise. You could then answer, "less than a millisecond."

 

[PeroK]

Look closely. The altitude used in my treatment is 105 meters, which is what the question specified. The difference between my method and the approximate method is that I've provided the classically exact solution. No approximation necessary. Naturally, the approximation in this case is very close to the truth. But the asker might find it convenient to have the exact solution for other cases, in which the length of the fall is a substantial fraction of (or is greater than) the radius of Earth.

The OP is four years old and the poster hasn't been seen again since. With any luck he has graduated by now.

 

[PeroK]

Well, you have neglected friction with the air, the rotation of the earth, the gravitational attraction from the moon, the non-spherical shape of the earth, and so on. So you have not provided an exact solution.

... not to mention the physical size of the bag!

 

[gmax137]

The OP is four years old and the poster hasn't been seen again since. With any luck he has graduated by now

ooops.

 

[Jenab2]

I neglected general relativity, too. But the point is that the solution that I found doesn't have any sources of error that the integrative approximation has, and indeed my solution has one fewer cause of error. And the one-and-done approximation from the high school physics textbook has, additionally, the error that the gravity field isn't constant across the fall, indeed the acceleration varies in that approximation more than it does in the integrative approximation method.

 

[Jenab2]

Well, you have neglected friction with the air, the rotation of the earth, the gravitational attraction from the moon, the non-spherical shape of the earth, and so on. So you have not provided an exact solution.

please do not confuse

with the "truth." All of the approximations you made would affect this result (leaving aside the question of significant digits: your answer assumes initial velocity is actually 5.000000000 m/sec and altitude is actually 105.0000000 meters).

If the question had been, "estimate the effect of assuming constant acceleration in determining the time to fall..." then your approach might be wise. You could then answer, "less than a millisecond."

Well, since this is a hypothetical problem, probably presented as someone's homework, then we may proceed as if the initial conditions are exact. I'll admit that I retain a ridiculous number of significant digits, but, as to that, I am merely reporting what my calculator told me. If this were a practical case in which it was important to know the probable error in the initial conditions, then I'd run through several scenarios in which those conditions varied and report a result with error bars. But this isn't a practical case. It is a hypothetical case.

 

[Jenab2]

Sure, but your method is way beyond what is needed, and likely well beyond the level of the student. That makes it inappropriate.

And if it had been appropriate I wouid be asking you or a moderator to edit your post. We do not provide complete solutions, at least not straightaway. Please start with just providing hints and pointing out errors. See the Forum Rules.

This is why I have been a latent member of the Physics Forums for most of the ten years I've been here. Every time I work out a difficult problem concisely, in detail, I'm jumped on. As someone else noted, the original post was made many years ago, and he probably graduated long ago. I'm not, therefore, helping him to cheat on his homework. For those who wonder about the time-to-fall-in-a-plunge-orbit problem, I aimed to satisfy their curiosity on the matter, nothing more.

 

[Jenab2]

That is irrelevant to the question in post #1, which is at a very modest altitude. We can take gravitational acceleration as the standard constant g.

Oh I quite agree. The high school approximation is quite good enough in this case. But the equation I presented is even more accurate, and remains good when the acceleration of gravity does vary considerably over the fall. It was, therefore, worth deriving, presenting, and showing how it works in an example case.