Dropping an object from a moving plane.

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In summary, the conversation is about finding the angle of sight theta for a rescue plane to release a capsule to land near a person in the water. The solution involves calculating the horizontal distance using the time it takes for the object to fall and the speed of the plane. The final answer is 30.5 degrees.
  • #1
opticaltempest
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I have to solve the following problem:

"A rescue plane is flying at a constant elevation of 1200m with a speed of 430 km/h toward a point directly over a person in the water. At what angle of sight [tex]\theta[/tex] should the pilot release a rescue capsule if it is land near the person in the water?"

The answer we are given is 57 degrees.

Here is the setup I think we have:
(we are not given a diagram)

http://img137.imageshack.us/img137/5049/setupsq1.jpg [Broken]Wouldn't the angle needed depend on the "horizontal distance" as labeled in the graph? In other words, if the angle is a set 57 degrees then the horizontal distance must be some fixed value.

We are not given the "horizontal distance" in the problem.

I imagine I could figure out that distance since I know the answer of 57 degrees, but if I wasn't given the answer there is no way to have solved this problem without knowing the horizontal distance - correct?
 
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  • #2
I think I figured out the solution. I am posting my solution.

The object falls at a constant rate so we can determine the time the object takes to fall to the ground:

[tex]
\begin{array}{l}
y_f = - \frac{1}{2}gt^2 \\
y_f = - 4.9t^2 \\
- 1200 = - 4.9t^2 \\
t = 15.642 \\
\end{array}
[/tex]

We can now find the distance the object moves horizontally using the time we found earlier and knowing that

[tex]
x_f = v_i \cos \left( \alpha \right)t
[/tex]

Alpha is zero since the object is dropped and falls straight down.

[tex]
x_f = 430\cos \left( 0 \right)t
[/tex]

In the next step I also have to convert 430km/h to m/s

[tex]
x_f = \left( {\frac{{430{\rm{km}}}}{{1{\rm{h}}}}\frac{{1{\rm{h}}}}{{3600{\rm{s}}}}\frac{{{\rm{1000m}}}}{{{\rm{1km}}}}} \right)\cos \left( 0 \right)t
[/tex]

[tex]
x_f = 119.44t
[/tex]

[tex]
x_f = 119.44\left( {15.642} \right)
[/tex]

[tex]
x_f = 1868.3{\rm{ meters}}
[/tex]Here are my final answers:

http://img164.imageshack.us/img164/7245/diagram0001np1.jpg [Broken]

So I guess the correct angle the problem was looking for was [tex]\gamma[/tex]. Does this solution look correct?

Except in my drawing [tex]\gamma[/tex] should be measured counterclockwise from the -90[tex]^{\circ}[/tex] axis (the arrow indicating the direction of measurement for [tex]\gamma[/tex] should be reversed.)
 
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  • #3
Actually, calculating that horizontal distance is precisely the problem. Once you know that, figuring out theta is trivial.

You can calculate how long it will take before the rescue capsule knocks the guy underwater. Since you also know the speed of the plane you can figure out at what horizontal distance you should release the capsule.

And by the way, I think the angle theta should be measured from the vertical.
 
  • #4
opticaltempest said:
We can now find the distance the object moves horizontally using the time we found earlier and knowing that

[tex]
x_f = v_i \cos \left( \alpha \right)t
[/tex]

Alpha is zero since the object is dropped and falls straight down.

[tex]
x_f = 430\cos \left( 0 \right)t
[/tex]

I don't get how the object falls straight down if it has a velocity equal to that of the plane..
 
  • #5
opticaltempest said:
[tex]
x_f = v_i \cos \left( \alpha \right)t
[/tex]

Alpha is zero since the object is dropped and falls straight down.
That's the only step that looks wrong. The horizontal velocity of the capsule is constant and equal to that of the plane. So that's why you use x=vt, where v is the velocity of the plane.
 
  • #6
here is the solution . i am not sure its right though because I didnt get the answer you mentioned.

430km/h=131m/s

s = 1/2gt^2
1200=5t^2
t=4*(15)^1/2 = 15.49


x = Vx(t)
x = 131(t)
x = 131*15.49= 2029.19

tan(0)= y/x
= 1200/2029.19= 0.59

therefore theta = 30.5

(edit : i might be a little off. I have used g=10m/s^2)
 

1. How does the speed of the plane affect the object's rate of fall?

The speed of the plane has no effect on the object's rate of fall. The object will fall at the same rate regardless of the speed of the plane.

2. Does the shape or weight of the object impact its rate of fall?

No, the shape or weight of the object does not impact its rate of fall. All objects, regardless of their shape or weight, will fall at the same rate due to gravity.

3. Will the object eventually reach a maximum speed while falling?

Yes, the object will reach a terminal velocity, which is the maximum speed it can reach while falling due to the resistance of air. This is usually around 120 miles per hour for most objects.

4. How does air resistance affect the object's fall?

Air resistance, also known as drag, slows down the object's fall. As the object falls, it pushes against the air molecules, causing the air to resist its movement. This resistance increases as the object's speed increases, eventually reaching a point where the drag force is equal to the force of gravity, resulting in a constant speed or terminal velocity.

5. Can the object's trajectory be affected by the plane's movement?

Yes, the object's trajectory can be affected by the plane's movement. If the plane is moving at an angle or changing direction, the object will also follow a curved path instead of falling straight down. However, this will not affect the object's rate of fall.

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