Aircraft question, speed and altitude and time interval?

Bianca Lannier: Parabolic Flight Path

Karl Samiec: During the ballistic flight, the aircraft moves under the sole influence of earth's gravity, just like a projectile. The components of its velocity when entering the zero-g flight arevx = 143 cos45 = 101 m/svy = 143 sin 45 = 101 m/sand at the end of the parabolic flight path arev'x = 143 cos(-45) = 101 m/sv'y = 143 sin (-45) =-101 m/sIn what follows I neglect the change of g in function of altitude and assume that it is always 9.8 m/s²(a) The velocity of the aircraft at the top of the maneuver is the same as its horzontal component vx, that is 101 m/s(b) The height it climbs: Î"y = (0 - vy²)/2a = - (101²)/2×(-9.8) = 520 mThe altitude of the top of the flight pathy = y0 + Î"y = 31,000 ft + 520 m = 9450 m + 520 m = 9,970 m(c) The time interval spent in zero-gÎ"t = Î"(vy)/a = (v'y - vy)/-g = -202/-9.8 Î"t = 20.6 s(d) I'm not sure about the upward acceleration of the motion in this phase.- If upward means vertical a! nd upward, then the trajectory of the aircraft would be a parabola (not a circle) with positive concavity. At the bottom of that trajectory vy = 0 and vx remains the same as it was. The answer would be v = vx = 101 m/s- If upward doesnt necessary mean vertical, but a is a centripetal acceleration, then the trajectory is indeed a circle, and the aircraft speed at any position on the circle (not just the bottom) would bev = sqrt(aR) = sqrt(0.8×9.8×4,300) = 180 m/s²...Show more