"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.

Question

"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.

anonymous · Answer

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anonymous · Answer

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 · Answer

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

raffle_snaffle · Answer

When you solve for displacement I would think you need to take the two velocities and subtract them from each other..

anonymous · Answer

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 · Answer

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 · Answer

do you have the answers? I would like to try and solve this one as well on my own time.

anonymous · Answer

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 · Answer

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 · Answer

and why use 9.04 sec?

raffle_snaffle · Answer

Is that question at me or fg?

anonymous · Answer

Anyone lol. This just bothers me o-o

raffle_snaffle · Answer

I will think it through when I have my physics book handy. lol

raffle_snaffle · Answer

I forgot most of the 2D kinematic equations. Lol

anonymous · Answer

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