Two airplanes leave an airport at the same
time. the velocity of the first airplane is
750 m/h at a heading of 51.3
◦
. the velocity
of the second is 620 m/h at a heading of 163◦
.
how far apart are they after 2.4 h?
answer in units of m.

For the given problems, both the planes are 1486 m apart after 2.4 h.

Explanation:

Let us consider two airplanes as A1 and A2. We know that the

The distance covered by airplane A1 is

Similarly, the distance covered by airplane A2 is

As the airplanes take off at an angle from the base, so they will be exhibiting X and Y components of displacements.

X components of both the airplanes

         =

Y components of both the airplanes

        =

The net displacement after 2.4 hours will be equal to the distance at which both planes are airplanes to each other at 2.4 hours.

Therefore, Net displacement is

So, both the planes are 1486 m apart from each other after 2.4 hours.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

trig 15.3 cos