How Far Would Something Fall (Neglecting Air Resistance) If It Fell For 12 Seconds?

Question

I am asking this to see how high a microgravity drop tower would have to be to allow an experiment to drop for 12 seconds. The drop tower would of course be vacuum-sealed to allow the highest quality microgravity (hence "neglecting air resistance"). How far would something have fallen if in a vacuum after 12 seconds of free-fall?

Answer 1

d2 = d1 + v1t + 1/2at^2

d1 = 0
v1 = 0

d2 = (1/2) at^2

d2 = (0.5)(9.8 m/s^2)(12s)(12s) = 706 m

Answer 2

Depends. The force of gravity is F = GmM/R^2 = ma; where m is the "something" mass, M >>>>> m is the mass of the attracting body (e.g., Earth), R is the distance between the centers of mass, and G is a universal constant ~ 6.6 X 10^-11. a is the acceleration, which determines how long something takes to fall a distance D.

Thus, a = GM/R^2 and D = 1/2 at^2; where D is the distance something would fall in t = 12 seconds. Find a and you can solve for D under any air free conditions of M and R.

If your attracting mass M is Earth and the distance R is the distance from the center to Earth's surface, then you can use a = g = 9.81 m/sec^2 or 32.2 ft/sec^2. But if your microgravity experiments need the exact value for a, then you might want to conduct a few experiments to find the precise g = a for your location.

For example, W = mg; where W is weight (in pounds or Newtons), m is mass, and g is what you are looking for if you need the exact g. Thus, g = W/m; so pick a mass m and weigh it on a precision scale several times. Calculate g each time and take the average for the precise g value rather than use the averages given earlier.

Answer 3

Using a kinematics equation:
h=.5at^2

Substitute for the acceleration of gravity and time:
h=.5 * 9.8 * 12^2

Solve:
h=705.6 meters

That would be a very tall tower!