Free Fall Exercises

Exercise One, Choosing a Suitable Free Fall Frame of Reference (Axis).- A stone is dropped from a balloon that is descending at a uniform rate of 12 m/s when it is 1000 m from ground. a) Calculate the velocity and position of the stone after 10 s and the time it takes the stone to hit the ground. b) Solve the same problem as for the case of a balloon rising at the given velocity.

Exercise Solution:
We must remember some of the kinematics formulas:

v = vi + at
y = yi + vit + ½ at2

vi and yi are the initial velocity and position respectively and a is the acceleration.

The first step is to choose a suitable free fall vertical frame of reference or axis, which could be pointing upward or downward. If we choose the direction downward, the initial velocity is then +12m/s and the acceleration of gravity is taken as +g = + 9.8m/s2 If we choose the upward direction, the initial velocity is taken as -12m/s and the acceleration as - 9.8m/s2

To answer question a) of exercise one, it seems convenient a vertical axis pointing downward with the origin at 1000 m over ground. Of course, a different reference axis and origin could be chosen. The result will be consistent with any inertial frame.

The data is then taken as: a=+9.8 m/s2, vi=+12m/s, yi=0, t=10s
v=12 m/s+9.8 m/s2·10s=110 m/s
y=12 m/s·10s+½·9.8m/s2·(10s)2=610 m or 390 m from ground.
The time to hit the ground is obtained doing
y = 1000 m in y = yi + vit + ½ at2:
1000 m = 12 m/s·t + ½ ·9,8m/s2·t2 or
0 = 4,9t2 + 12t - 1000, t in seconds
Solving this quadratic equation results t = 13.1 s

To answer question b) of exercise one, we could maintain the same free fall frame of reference as before and then the velocity vi will be -12m/s and the same acceleration. Or, we can choose an upward direction axis with vi = + 12m/s and a = -9.8 m/s2.

The usual choice is this last one, i.e.
vi=+12 m/s, a=-9.8 m/s2, yi = 0, t = 10 s
v = 12 m/s - 9.8 m/s2·10s = -86 m/s
y = 12 m/s·10s - ½ ·9.8m/s2·(10s)2 = -370 m or 630 m from ground.

The time to hit the ground is obtained doing
y = -1000 m in y = yi + vit + ½ at2:
(*) -1000 m = 12 m/s·t - ½ ·9,8m/s2·t2 or
0 = 4,9t2 - 12t - 1000, t in seconds.

Solving this quadratic equation results t = 15.56 s

As a curiosity, if we had maintained the same free fall frame, this time it should be the same. Let's see:
+1000 m = -12 m/s·t + ½ ·9,8m/s2·t2 , which is the same equation (*) multiplied by -1 or 0 = 4,9t2 - 12t - 1000, t in seconds.
Then t = 15.56 s as before.

Other Exercises:
Go To Free Fall Exercises, Part Two
Go To Free Fall Exercises, Part Three

To See the Free Fall Physics Theory:
Go To Free Fall Theory

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