- #1
v3ra
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An airplane has an air speed of 4.20 × 10^2 km/h [N 45° E]. The wind speed is 30 km/h
to the west.
a) What is the airplane’s resultant velocity?
I decided to use the trigonometric solution, but I am confused on how to break the equation down in order to get 399 km/h
c2 = a2 + b2 - 2ab.cosC
c2 = (4.20 x 10^2)^2 + (30)^2 - 2( 420) x (30)cos135
c2 = (17640000 + 900) - 17819
c2 = 17623081
c2 = ? (I tried square root of 17623081, but of course this is far off from the answer)
Please assist!
to the west.
a) What is the airplane’s resultant velocity?
I decided to use the trigonometric solution, but I am confused on how to break the equation down in order to get 399 km/h
c2 = a2 + b2 - 2ab.cosC
c2 = (4.20 x 10^2)^2 + (30)^2 - 2( 420) x (30)cos135
c2 = (17640000 + 900) - 17819
c2 = 17623081
c2 = ? (I tried square root of 17623081, but of course this is far off from the answer)
Please assist!