In zero-gravity astronaut training and equipment testing, NASA flies a KC135A aircraft along a parabolic flight path. The aircraft climbs from 24 000 ft to 31 000 ft, where it enters the zero-g parabola with a velocity of 143 m/s nose high at 45.0° and exists with velocity 143 m/s at 45.0° nose low. During this portion of the flight, the aircraft and objects inside its padded cabin are in free fall; they have gone ballistic. The aircraft then pulls out of the dive with an upward acceleration of 0.800g moving in a vertical circle with radius 4.13 km. (During this portion of the flight, occupants of the aircraft perceive an acceleration of 1.8g) What are the aircraft's (a) speed and (b) altitude at the top of the maneuver? (c) What is the time interval spent in zero gravity? (d) What is the speed of the aircraft at the bottom of the flight path?
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!
2 answers