Projectile Motion? (10 points)

At 2:00PM the Nimitz is 450 KM Southeast of a docked ACC. The Nimitz is traveling due west at 60.2 KPH. At 4 PM a plane takes off from the docked carrier and must land on the Nimitz at 6:30. There is a 40 km wind from the Southwest to the Northeast. In what direction an at what speed relative to the air mush the pilot fly?

Show all work in detail and include a diagram and description of what the numbers represent.

Let's draw a diagram. Here's the initial position, at 2:00. (E represents East, S is South, C is the carrier, and N is the Nimitz. Since the Nimitz is southeast of the carrier at 2:00, the angle is 45°.)

C------------------ E
.|..\ 45°
.|......\
.|..........\ 450 km
.|..............\
.|..................\
S.....................N

So, at 6:30, when the plane will land, the Nimitz will have moved west. Traveling at 60.2 km/hr for (6:30 - 2:00) = 4.5 hr, the westward motion will be 60.2 km/hr * 4.5 hr = 270.9 km

C------------------ E
.||.\ 45°
.|.|....\
.|..|.......\ 450 km
.|...|A........\
.|....|.............\
.|....N----------------\
S.......270.9 km

So, the new vector from the carrier to the Nimitz, A, is the sum of 450 km SE and 270.9 km W. We'll convert the SE vector to E and S components, to be able to add the vectors. The E component of the SE vector is

450 km * cos 45° = 450 km * √2 / 2 = 225 km * √2

The S component of the SE vector is

450 km * sin 45° = 450 km * √2 / 2 = 225 km * √2

So, when we add the two vectors together (remember that west is the same as negative east), we get the vector from the carrier to the Nimitz at 6:30:

A = (225 km * √2 - 270.9 km) * E + 225 km * √2 S

Now, the plane needs to travel this vector between 4:00 and 6:30. The vector the plane flies will be the sum of the motion of the plane traveling with no wind, P, plus the distance the wind carries the plane, W.