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?
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
To calculate the constant acceleration required to stop an experiment package in the remaining distance after falling for 5.28 seconds, we can use the kinematic equation for uniform acceleration:
d = vit + (1/2)at^2
where:
- d is the distance traveled
- vi is the initial velocity
- a is the acceleration
- t is the time
In this case, the distance traveled is the height of the shaft minus the remaining distance, which is given as 156 m. The initial velocity, vi, is 0 since the object is dropped. The time, t, is given as 5.28 seconds.
Let's plug in the values into the equation:
156 = 0 + (1/2)a(5.28^2)
Simplifying the equation:
156 = (1/2)(27.8784)a
Dividing both sides by 13.9392:
11.2 = a
Therefore, the constant acceleration required to stop the experiment package in the remaining distance is 11.2 m/s².