anonymous
  • anonymous
"2.5D motion" An airplane must drop supplies on a target, vertical distance = 400m, horizontal distance = 3500m, initial velocity = 250 m/s East. If the wind blows toward the South at 50 m/s, find the horizontal displacement while the package is in the air.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1378596237362:dw|
anonymous
  • anonymous
So first, I solved for the time for the package to hit the ground. dy = 0.5at^2 400 = 0.5(-9.8)t^2 t=9.04 seconds
anonymous
  • anonymous
Then, the time the pilot needs to wait to drop supplies... dx = vit 3500 = 250t t=14 sec 14sec-9.04sec = 4.96sec before release

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raffle_snaffle
  • raffle_snaffle
When you solve for displacement I would think you need to take the two velocities and subtract them from each other..
anonymous
  • anonymous
Then, the final velocity of the package: Vfy: vf^2 = vi^2 + 2ad vf^2 = 0^2 + 2ad vf^2 = 2(-9.8)(400) vf = 88.6m/s downward Vfx: vf=vi=250 m/s East Resolve: sqrt(88.6^2+250^2) = 265m/s Theta=tan^-1(88.6/250) = 19.5 degrees 265 m/s @ 19.5 degrees below horizontal
anonymous
  • anonymous
So, If the wind blows toward the South at 50 m/s, find the horizontal displacement while the package is in the air.
raffle_snaffle
  • raffle_snaffle
do you have the answers? I would like to try and solve this one as well on my own time.
anonymous
  • anonymous
In the first equation you assumed that the initial velocity in the "y" direction was 0m/s, but doesn't the package has an initial velocity in the "y" axis? That velocity would be the 50m/s of the wind blowing south.
anonymous
  • anonymous
The package has 0m/s in the downward direction to begin with, but the South is part D in the problem. South and y-velocity aren't the same because it's a different dimension. The solution in the notes: 3500-1240=2260m East d South: d=vt d=50(9.04) d=452 South Theta= tan^-1(452/2260) R=sqrt(452^2+2260^2) = 2305m @11.3 degrees South of East I just don't know where the 1240 came from.
anonymous
  • anonymous
and why use 9.04 sec?
raffle_snaffle
  • raffle_snaffle
Is that question at me or fg?
anonymous
  • anonymous
Anyone lol. This just bothers me o-o
raffle_snaffle
  • raffle_snaffle
I will think it through when I have my physics book handy. lol
raffle_snaffle
  • raffle_snaffle
I forgot most of the 2D kinematic equations. Lol
anonymous
  • anonymous
|dw:1378596995254:dw|
anonymous
  • anonymous
This is how it's supposedly visuallized
raffle_snaffle
  • raffle_snaffle
Why are you using the 11.3*?
anonymous
  • anonymous
11.3 is the resultant angle of the distances South and East
anonymous
  • anonymous
It had to be solved for
raffle_snaffle
  • raffle_snaffle
So you are assuming that is the angle you need to use?
anonymous
  • anonymous
No, it was solved for after the distances East and South were solved for in part D. The problem is I don't know: 3500-1240=2260m East where did 1240 come from. and why in: d=vt d=50(9.04) d=452 South 9.04 was used
anonymous
  • anonymous
I think that horizontal displacement in East direction, is only caused by velocities in East direction, being that speed independent from the speed of the wind. So: D(East)=V(East) * TimeInTheAir D=250m/s * 9.04s D=2260m East direction
anonymous
  • anonymous
Oh, but they ask you for total displacement in the horizontal plane. In that case, you could use the magnitude of the resultant velocity (V=254.95m/s) and multiply it by Time=9.04s to get the total displacement in the resultant direction. D=V*TimeInTheAir D=254.95m/s * 9.04s D=2305m in the resultant velocity direction
raffle_snaffle
  • raffle_snaffle
Why are you solving for the time the pilot needs to drop the supplies?
raffle_snaffle
  • raffle_snaffle
The speed of the pilot and package have the same velocity. So you don't need to calculate the time the pilot needs to drop the supplies.
raffle_snaffle
  • raffle_snaffle
I meant supplies
anonymous
  • anonymous
That was part B
raffle_snaffle
  • raffle_snaffle
I am trying to work out the problem. I thought you were asking why d=vt =50m/s(9.04s) you use the 9.04s because the supplies hit the the ground at that time.
raffle_snaffle
  • raffle_snaffle
Would you agree?
raffle_snaffle
  • raffle_snaffle
This is how I solved it. I made the problem more complicated than it really needs to be. Lol|dw:1378605321842:dw| find the third vecctor. |dw:1378605405319:dw| now |dw:1378605508330:dw|
anonymous
  • anonymous
Oh yeah okay I understand now. Thank you both for your help.

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