At a NASA research center free-fall research is performed by dropping experiment

Question

At 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 is the maximum time interval for free fall if an experiment package were to fall the entire 156 m? Actual NASA specifications allow for a 5.19 s drop time interval. How far do the packages drop in 5.19 s? What is their speed at 5.19 s? NASA does not want the experiment package to slam into the ground and get smashed to pieces. They fill the bottom of the shaft with lightweight foam to break the fall. What constant acceleration would be required to gently stop an experiment package in the distance remaining in the shaft after its 5.19 s fall? Draw position, velocity, and acceleration graphs. These must be turned in on paper.

Explanation / Answer

Given that,

height,hi = 156 m

Universal earth's gravity,
g = 9.81 m/s^2

(a) Maximum time of fall

Tf = sqrt ( [2h] / g )
Tf = sqrt ( [ 2 * (156 m) ] / (9.81 m/s^2))
Tf = 5.639 s

(b) Distance fallen is 5.19 s

hf = 0.5 * g * (Tf)^2
hf= 0.5 * (9.81 m/s^2) * (5.19 s)^2
hf = 132.12 m

(c) Velocity after 5.19 s

Vf = sqrt{ Vi^2 + [2gd] }
Vf = sqrt { (0.0 m/s)^2 + [ 2 * (9.81 m/s^2) * (132.12 m) ] }
Vf = sqrt { 2592.2 m^2/s^2 }
Vf = 50.91 m/s

(d) Acceleration is,
a = [ Vf^2 - Vi^2 ] / [2d]

d=hi-hf=156 m-132.12 m=23.88 m
a = [ -(50.91 m/s)^2 ] / [ 2 * (23.88 m) ]
a = [ -2592 m^2/s^2 ] / [ 47.76 m ]
a = -54.27 m/s^2

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