In zero-gravity astronaut training and equipment testing, NASAflies a KC135A air
ID: 1759544 • Letter: I
Question
In zero-gravity astronaut training and equipment testing, NASAflies a KC135A aircraft along a parabolic flight path. As shown inthe figure, the aircraft climbs from 24,000 ft to 37000 ft, where it enters the zero-g parabolawith a velocity of 119 m/s at 45.0°nose high and exits with velocity 119 m/sat 45.0° nose low. During this portion of the flight, theaircraft and objects inside its padded cabin are in free-fall -they have gone ballistic. The aircraft then pulls out of the divewith an upward acceleration of 0.800g, moving in avertical circle with radius 4.22 km.(During this portion of the flight, occupants of the plane perceivean acceleration of 1.800g.)(a) What is the aircraft's speed at the top of the maneuver?
(c) What is the time spent in zero gravity?
(d) What is the speed of the aircraft at the bottom of the flightpath?
Explanation / Answer
(a)
according to the given problem
from the theory of kinematics we get the speed at thetop as
vx = vi cosi
= (119 m / s)(cos45o)
= ....... m / s
(b)
the equation of motion is given by
vyf2 =vyi2 + 2 ay (yf -yi)
0 = ((119 m / s) (cos45o))2 + 2(-9.8 m / s2) (yf - 37000 ft)
here we use the conversion (3.28 ft = 1 m)
so the altitude at the top of the maneuver will be
yf = (37000 ft) + (4220 m) (3.28 ft / 1m)
= ........ ft
(c)
in case of zero gravity for the free fall motion weget
vfy = vyi + ay t
- vx = + vx - (9.80 m /s2) t
t = ...... s
(d)
at the bottom of the flight path the accelerationis
ac = v2 / r
v = (ac r)
= [(0.8) (9.80 m /s2) (4130 m)]
= ........ m / s