### Video Transcript

The F-35B is a short takeoff and
vertical landing fighter jet. An F-35B takes off vertically
upward to a height of 20.00 meters, while still facing horizontally. The fighter then follows a flight
path that is angled at 30.00 degrees above a line parallel to the ground, 20.00
meters vertically above ground. The fighter flies for a distance of
20.00 kilometers along this trajectory. What is the fighterβs final
displacement?

We can call the displacement of the
fighter after these maneuvers π, where we understand that π is a vector with both
magnitude and direction. To begin on our solution, letβs
draw out a sketch of this F-35Bβs motion. Weβre told that the jet fighter
rises vertically a distance weβve called β of 20.00 meters. And then begins to move at an angle
of 30.00 degrees to the horizontal along a path of length 20.00 kilometers. The fighterβs displacement π is
the vector that connects its initial location on the ground with its final location
at the end of its 20.00 kilometer of leg. Itβs that vector that we want to
solve for based on this given information. We can start by orienting the
motion of our fighter to a set of coordinate axes. Weβll position a pair of π₯- and
π¦-coordinate axes, so that their origin, the place where they cross, is the place
where our jet fighter began its journey. This means we can write our
displacement vector π in the following form. A component, weβll call it π sub
π₯, in the π-direction plus a component, weβll call it π sub π¦, in the
π-direction.

So now, our task is to set out and
solve for π sub π₯ and π sub π¦. Looking first at π sub π₯, we see
that that value, the π₯-component of the jet fighterβs motion, will be equal to π
times the cos of 30.00 degrees. We know that π is given as 20.00
kilometers or 20.00 times 10 to the third meters. When we multiply that by the cos of
30.00 degrees, we find a value of 1.732 times 10 to the fourth meters. Thatβs π sub π₯. Now we want to solve for π sub π¦,
the vertical displacement of the jet fighter. Looking at our diagram, we see that
π sub π¦ is equal to β plus π times the sin of 30.00 degrees. Plugging in for β and π, we once
again use a value of 20.00 times 10 to the third meters for π. And β, weβre given as 20.00
meters. With these values entered on our
calculator, we find that π sub π¦, to four significant figures, is 1.002 times 10
to the fourth meters. We insert this value for π sub π¦
in our expression for displacement, π. And this gives us the final form
for the displacement of the jet fighter, 1.732 times 10 to the fourth meters in the
π-direction and 1.002 times 10 to the fourth meters in the π-direction.