Dropping an object from a moving plane.


by opticaltempest
Tags: dropping, moving, object, plane
opticaltempest
opticaltempest is offline
#1
Sep20-06, 03:18 PM
P: 135
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)




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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opticaltempest
opticaltempest is offline
#2
Sep20-06, 03:38 PM
P: 135
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}}}}\f rac{{{\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:



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.)
Galileo
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#3
Sep20-06, 03:45 PM
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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.

QuantumKing
QuantumKing is offline
#4
Sep20-06, 05:24 PM
P: 46

Dropping an object from a moving plane.


Quote Quote by opticaltempest
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 dont get how the object falls straight down if it has a velocity equal to that of the plane..
Galileo
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#5
Sep21-06, 12:09 AM
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Quote Quote by opticaltempest
[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.
konichiwa2x
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#6
Sep21-06, 01:27 AM
P: 81
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)


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