# Plane delivering supplies

1. Jul 18, 2013

### Zondrina

1. The problem statement, all variables and given/known data

A plane is attempting to deliver emergency food and supplies to a community.

The plane has a horizontal velocity of 140 km/h as it flies at an altitude of 180 m.

The community is situated on a patch of land which is 72m x 72m = 5184 m2.

a) If the supplies are released as the plane flies directly overhead, will they touch down on land or fall into the water?

b) When should the supplies be released so they can touch down very close to the center of the patch of land. Answer in terms of distance to the target, not time.

2. Relevant equations

$\vec{v}_H = 140 km/h = 38.9 m/s [Forward]$
$\vec{Δd}_V = 180 m [Up]$ <- I'm interpreting the constant altitude as vertical displacement from the ground.

3. The attempt at a solution

a) I believe the first thing I want to do is find out how long the supplies will be dropping for. I can do that by using :

$\vec{Δd}_V = v_1 Δt - (1/2) \vec{a} (Δt)^2$

There is no initial velocity, and I switch the direction of the acceleration from [Down] to [Up] by using a negative sign. Subbing everything in I get :

$180 = 0 - (4.9)(Δt)^2$
$180 = -(4.9)(Δt)^2$

I seem to be having an issue there ^, can't factor that in the reals. If I'm wrong about the acceleration, why?

After I get that figured, I can use the newly found time to compute $\vec{Δd}_H = \vec{v}_H Δt$ which would be the horizontal displacement I'm looking for.

Using the horizontal displacement, if it it's bigger than 72m, then the supplies unfortunately got lost in the water. If it's smaller than 72m they landed safely in the patch of land.

It's just the arithmetic I need a bit of help with, thanks.

b) The center of the dry patch of land is 36m in. I would need to know how much the supplies missed by in part a) I guess before I could continue this.

2. Jul 18, 2013

### CAF123

You took the negative direction to be pointing downwards. Recall that displacement is also a vector quantity.

3. Jul 18, 2013

### Zondrina

a) So I should be considering $\vec{Δd}_V$ to be [Down] not [Up] so I can consider [Down] the positive direction. Then the acceleration is also in the positive direction.

Then I get : $180 = (4.9)Δt$ which works out to $Δt = 6.06 s$.

So $\vec{Δd}_H = \vec{v}_H Δt = (38.9)(6.06) = 235.7 m [Forward]$ will be the distance that the aide travels horizontally. Therefore the supplies completely missed the patch of land by $235.7m - 72m = 163.7 m$ and fell into the water instead.

b) The center of the dry patch of land is 36m in. We want to get as close to dropping the supplies there as possible.

If the pilot dropped the supplies 163.7 m before the patch of land, then the supplies would land right on the border of the community. So to hit the center, he would need to drop the supplies $164.7m - 36m = 128.7m$ before he reached the community to insure they would be as close to the center of the town as possible.

4. Jul 18, 2013

### CAF123

Or, to keep with your original set up, with negative downwards and the x axis along the horizontal trajectory of the plane means that the object is displaced -180 m so that your equation becomes -180 = -4.9t2.