Answer:
The true course:
north of east
The ground speed of the plane: 96.68 m/s
Explanation:
Given:
= velocity of wind = 
= velocity of plane in still air = 
Assume:
= resultant velocity of the plane
= direction of the plane with the east
Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

Let us find the direction of this resultant velocity with respect to east direction:

This means the the true course of the plane is in the direction of
north of east.
The ground speed will be the magnitude of the resultant velocity of the plane.

Hence, the ground speed of the plane is 96.68 km/h.
4. The Coyote has an initial position vector of
.
4a. The Coyote has an initial velocity vector of
. His position at time
is given by the vector

where
is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with


The Coyote hits the ground when
:

4b. Here we evaluate
at the time found in (4a).

5. The shell has initial position vector
, and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is

We find the shell hits the ground at

5b. The horizontal component of the bullet's position vector is

where
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:

Answer: f=20 (i think)
Explanation:
all I did was divide 300 and 15.
300/15= 20
Answer:
radius comes out to be 3 m
height of the cylinder comes out to be 3m
Explanation:
given
volume of cylinder = 27π m³
π r² h = 27π
r² h = 27.............(1)
surface area of cylinder open at the top
S = 2πrh + π r²




for least amount of material requirement.

hence radius comes out to be 3 m
for height put the value in the equation 1
so, height of the cylinder comes out to be 3m
Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .