how far did it fall in 5.28 s?
d = 4.9 t^2 = 4.9 (5.28)^2
how fast is it going?
Vi = g t = 9.81 * 5.28
now how much shaft is left?
y = 156 - d
so we have to stop it from Vi to v = 0 in y meters
0 = Vi - a t
so t = Vi/a = time to stop
y = Vi t - .5 a t^2
y = Vi (Vi/a) - .5 Vi^2/a
y = .5 Vi^2/a
or a = .5 Vi^2/y
t a NASA research center free-fall research is performed by dropping experiment packages from the top of an evacuated shaft 156 m high. Free-fall imitates the so-called microgravity environment of a satellite in orbit.
What constant acceleration would be required to stop an experiment package in the distance remaining in the shaft after its 5.28 s fall?
1 answer
1 answer