Optimal Angle for Westward Flight with Crosswind | Vector Problem Solution

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To fly directly west with a crosswind from the south at 30 m/s, the pilot must adjust the heading to account for the wind. The engine speed is 100 m/s, and using trigonometry, the angle can be calculated by taking the tangent inverse of the wind speed divided by the plane's speed, resulting in an angle of 17 degrees. This angle must be added to the westward direction (270 degrees), leading to a final heading of 287 degrees. The discussion emphasizes the importance of using correct trigonometric relationships to find the optimal flight angle. Proper calculations ensure the aircraft maintains a straight path despite the crosswind.
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Homework Statement


A pilot wants to fly directly to the west. The engine pushes the plane at 100 m/s, and there is a crosswind blowing to the south at 30 m/s. Determine the exact angle at which the pilot should head.



Homework Equations





The Attempt at a Solution


Here's what I did:

100^2+30^2= 10900

I then did SqRT of 10900 and got 104

then I did tan negative 1 to find the angle and got 89 degreess

right?
 
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Hmm, I got 287 degrees. Check your work again.
 
my work shows how I got 89

how did you get your answer?
 
Well I did it using vectors and then by using a "whiz wheel" and both answers came out the same. You got 104 which is correct, but you can't just take the tan-1 of that. You have to take tan-1 of .3 because 30/100 = .3. That gives you 17 degrees but now if you are heading west and the wind is out of the north you have to point the aircraft into the wind to crab, so you will still track on a westerly heading. Therefore 270 + 17 = 287 degrees.
 
why 30/100 and not 30/100000, 30/89, 30/76, get what I am saying?
 
Simple trigonometry..

Opposite side = 30
Adjacent side = 100

Tan theta = opp divided by adjacent

To find the angle, you'd take the tangent inverse of it.

Tan-1 = 30/100

Which produces 17
 
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Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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