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April 1, 2015

April 1, 2015

Posted by **Ania** on Thursday, October 18, 2012 at 6:22pm.

i have done part a like this :

given:

(at begining of parabolic flight)

vx = 143 cos45 = 101 m/s

vy = 143 sin 45 = 101 m/s

(at end of parabolic flight)

v'x = 143 cos(-45) = 101 m/s

v'y = 143 sin (-45) =-101 m/s

The velocity of the aircraft at the top of the maneuver is the same as its horzontal component vx, that is 101 m/s

i don't know how to do the rest and i'm not even sure if part a) is right. please help! and please explain reasoning and why you chose certain equations and stuff. thank you!

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