A flag pole is attached to the side of a building 5.00 ft off the ground and it hangs down 30 degrees from the vertical. If the pole is 6.00 ft in length, how high in feet is the top of the pole at its highest point (sticking out from building)? (Hint: draw a picture) E

   h ’= 0.6963 m

this is the height of the part of the post it supports.

Explanation:

For this exercise we can use the trigonometric relationships

Let's start by finding the vertical component of the height of the pole

             cos 30 = x / L

             x = L cos 30

             x = 6 cps 30

             x = 5.196 m

with this length, the post has the same height as the building, so there are

               Aₓ = L -x

               Δx = 6 -5.196

               Δx = 0.804 ft

the height of this section is

               cos 30 = h ’/ Dx

               h ’= 0.804 cos30

               h ’= 0.6963 m

this is the height of the part of the post it supports.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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we need to know the time interval during which the force acts. impulse of force = momentum change