Rescue Plane: Release Angle of 57.3°

[stupif]

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

 

[AJKing]

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).

$h = 1.2km = 1200m$

$v = 430km/h = 430000m/3600s \approx 120m/s$

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.

 

[stupif]

what is the speed i should take?

 

[AJKing]

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:

$KMS$

$Kilogram, Meter, Second.$


If this helps you remember:

$Kiss, My, Sock$


Always use these units.

 

[stupif]

no～i mean i should take vertical speed or horizontal speed?

 

[AJKing]

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.

 

[AJKing]

Can I see your work on this question so far?

 

[stupif]

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~

 

[rl.bhat]

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

 

[stupif]

how to find the vertical velocity?

 

[stupif]

thank you, i got it...
thank you very much...