Rescue Plane: Release Angle of 57.3°

In summary: Use the following equation to calculate the velocity of the parcel when it is released from the plane: v = (h – tan θ) / (2 * π * t).In summary, the pilot needs to release the crate at an angle of 57.3 degrees from the vectical in order to hit the target in the water.
  • #1
stupif
99
1
1. a rescue plane is flying at a constant elevation of 1.2km with a speed of 430km/h toward a point directly over a person struggling in the water. at what angle of sight should the pilot release a rescue capsule if it is to strike the person in the water...answer is 57.3degreen from vectical.



2. i don't know what the question talking about. can someone explain to me( with diagram is better)?? thank you



The Attempt at a Solution

 
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  • #2
I'd make you a diagram if I had an art program on this computer, but alas . . . here are your variables. Always remember to write your variables. It makes visualization so much easier (for me, at least).

[itex]h = 1.2km = 1200m[/itex]

[itex]v = 430km/h = 430000m/3600s \approx 120m/s[/itex]

And the question is basically asking:

A pilot is flying a plane 1200m above the ground, at a speed of 120m/s, and in front of him is a target. The pilot is tasked with dropping a supply crate directly on that target without slowing down or lowering his altitude. What angle does the pilot need to be at (in relation to the target) when he releases the crate so that it hits the mark dead on?



Keep in mind that the crate is initially moving at the same velocity as the plane, so it will carry forward when it's released. That's why there needs to be an angle of difference between the mark and the plane.

Let me know if you're with me still.
 
Last edited:
  • #3
what is the speed i should take?
 
  • #4
stupif said:
what is the speed i should take?

Use the speed 120m/s, but just know that I rounded that speed up from 119.444m/s.

Remember this FOREVER:

[itex]KMS[/itex]

[itex]Kilogram, Meter, Second.[/itex]


If this helps you remember:

[itex]Kiss, My, Sock[/itex]


Always use these units.
 
  • #5
no~i mean i should take vertical speed or horizontal speed?
 
  • #6
stupif said:
no~i mean i should take vertical speed or horizontal speed?

You need to investigate both.

Imagine dropping a ball while standing still and think about how it moves. Compare that to the motion of a ball dropped while running.

If you drop a ball while standing still, the only noticeable force acting on it is the gravitational acceleration.

And, if you drop a ball while running, the ball's path is again effected by gravity (like before), but now also your horizontal velocity.

In the second example here, both horizontal and vertical velocities change the position of the ball. Just like with the crate dropped from the plane.
 
  • #7
Can I see your work on this question so far?
 
  • #8
okay
1st i find vertical speed. vertical speed = 119.44/tan angle
2nd i find angle, i use this formula, v^2= u^2+2as, 0= (119.44/tan angle)^2 +2(9.8)(1200)
angle= 37.91degree from vertical
and the answer is wrong~
 
  • #9
Find the time t taken by the parcel to cover 1200 m height to reach the ground. Take initial vertical velocity of the parcel zero.
Find the horizontal distance covered by the parcel before it hits the person, using x = horizontal velocity x time.
Then tanθ = x/h
 
  • #10
how to find the vertical velocity?
 
  • #11
thank you, i got it...
thank you very much...
 

Related to Rescue Plane: Release Angle of 57.3°

1. What is the significance of the release angle of 57.3° in a rescue plane?

The release angle of 57.3° is the optimal angle for a rescue plane to release a life-saving package or equipment. This angle ensures that the package reaches the intended target with maximum accuracy and efficiency.

2. How is the release angle of 57.3° determined for a rescue plane?

The release angle of 57.3° is determined through extensive calculations and simulations by scientists and engineers. Factors such as wind speed, altitude, and weight of the package are taken into account to determine the optimal angle for release.

3. Why is the release angle of 57.3° considered important in rescue operations?

The release angle of 57.3° is crucial in rescue operations as it ensures that the package or equipment reaches the intended target accurately and efficiently. This is especially important in emergency situations where every second counts and any errors in release angle could result in failure to reach the target.

4. Can the release angle of 57.3° be adjusted based on different factors?

Yes, the release angle of 57.3° can be adjusted based on various factors such as wind speed, altitude, and weight of the package. Scientists and engineers continuously monitor these factors and adjust the release angle accordingly to ensure the success of the rescue operation.

5. Are there any risks associated with using the release angle of 57.3° in a rescue plane?

The release angle of 57.3° is determined to be the optimal angle for rescue operations and is extensively tested and calculated. However, there may still be some risks associated with using this angle, such as unexpected changes in weather conditions. It is important for rescue teams to continuously monitor and adjust the release angle to ensure the safety and success of the operation.

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