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 153 m high. Free-fall imitates the so-called microgravity environment of a satellite in orbit. Hint: In this problem, all the motion is downward, so make the top y=0, and have the positive y-axis pointing downward.

(a) What is the maximum time interval for free fall if an experiment package were to fall the entire 153 m?

(b) Actual NASA specifications allow for a 5.19 s drop time interval. How far do the packages drop in 5.19 s?

(c) What is their speed at 5.19 s?

(d) 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?
  ---Select--- upward downward

Explanation / Answer

a) let time taken is t seconds.

then using the formula,

distance covered=initial velocity*time+0.5*acceleration*time^2

and noting that initial veloicty=0 and acceleration=9.8 m/s^2

we get

153=0*t+0.5*9.8*t^2

==> t=5.588 seconds

b)distance travelled in 5.19 seconds=0.5*9.8*5.19^2=132 m

hence the package will drop 132 m in 5.19 seconds.

c) speed at 5.19 seconds=initial veloicty+acceleration*time

=0+9.8*5.19=50.862 m/s

d)let acceleration required be a.

this acceleration will be upwards so as to slow down the package.

initial veloicty=50.862 m/s

distance to be covered=153-132=21 m

final veloicty=0

using the formula:

final veloicty^2-initial velocity^2=2*acceleration*distance

0^2-50.862^2=2*a*21

==> a=-61.594 m/s^2

negative sign shows that the acceleration should be in upwards direction.

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