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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
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
When you solve for displacement I would think you need to take the two velocities and subtract them from each other..
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
So, If the wind blows toward the South at 50 m/s, find the horizontal displacement while the package is in the air.
do you have the answers? I would like to try and solve this one as well on my own time.
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.
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.
and why use 9.04 sec?
Is that question at me or fg?
Anyone lol. This just bothers me o-o
I will think it through when I have my physics book handy. lol
I forgot most of the 2D kinematic equations. Lol
This is how it's supposedly visuallized
Why are you using the 11.3*?
11.3 is the resultant angle of the distances South and East
It had to be solved for
So you are assuming that is the angle you need to use?
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
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
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
Why are you solving for the time the pilot needs to drop the supplies?
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.
I meant supplies
That was part B
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.
Would you agree?
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|
Oh yeah okay I understand now. Thank you both for your help.