Calculating Flight Direction & Velocity: Bush Pilot Flying 250.0 km [N30.0°E]

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To reach the lake located 250.0 km at a bearing of N30.0°E, the bush pilot must adjust her heading to account for the wind blowing from the west at 40.0 km/h. Calculations for the correct heading involve using trigonometric functions, but initial attempts have led to confusion regarding the angle adjustment needed for the specific direction. The pilot's airspeed of 210.0 km/h must be combined with the wind speed to determine her ground velocity. A diagram may help visualize the problem and clarify the relationships between airspeed, wind speed, and resultant ground velocity. Accurate calculations are essential for successful navigation to the destination.
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A bush pilot wants to fly her plane to a lake that is 250.0 km [N30.0°E] from her starting point. The plane has an air speed of 210.0 km/h and a wind is blowing from the west at 40.0 km/h.

a) In what direction should she head the plane to fly in directly to the lake?
After determining Vpa, Vag I tried heading = sin^-1 (40/210) = 20, but i think that would only work if it was just 250.0 km instead on 250.0 km [N30.0°E].



b) If she uses the heading determined in (a), what will be her velocity relative to the ground?
no idea since i can't get a.
 
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Why don't you draw a diagram?
 
There is my diagram but I'm not sure if it is right.
 

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

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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