Finding Vector Components using Trigonometry

AI Thread Summary
To find the X and Y components of a vector using trigonometry, the sine and cosine functions should be applied correctly. The hypotenuse of the triangle is given as 147 km, with the angle being 71.2 degrees. The vertical component (Y) is calculated using sine, resulting in Fy = sin(71.2) * 147, while the horizontal component (X) uses cosine, yielding Fx = cos(71.2) * 147. The calculated components are Fx = 47.4 km and Fy = 139 km, which are confirmed to be correct. Proper application of trigonometric functions is essential for accurate vector component calculations.
C42711
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By c42711 at 2011-10-30

I am supposed to find the X and Y components of the above vector. What I tried to do first was dot in a line, forming a triangle. I then tried to find the length of that line using tangent. I got 431.8, but I am not sure that is right. Next, I used the Pythagorean theorem to try to find the X and Y components, I got some crazy numbers, so I know I'm doing something wrong. Any help would be fantastic. Please & thank you.:)
 
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Tangent is not the correct function to use. 147km is the hypotenuse of the triangle.
 
You'd need to use sine for the vertical and cosine for the horizontal. The sine of the angle is y / the hypotenuse, so y = sin71.2 * h. Similarly, x = cos71.2 * h. Hope that helps.
 
So I tried that and got:
Fx= 47.4 km
Fy= 139 km

Does that seem correct?
 
Yeah, those are the correct values.
 
Than you so much, you're the best.:)
 
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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