**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 = v_{i} + at

y = y_{i} + v_{i}t + ½ at^{2}

v_{i} and y_{i} 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/s^{2}
If we choose the upward direction, the initial velocity is taken as -12m/s
and the acceleration as - 9.8m/s^{2}

__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/s^{2},
v_{i}=+12m/s, y_{i}=0, t=10s

v=12 m/s+9.8 m/s^{2}·10s=110 m/s

y=12 m/s·10s+½·9.8m/s^{2}·(10s)^{2}=610 m or 390 m from ground.

The time to hit the ground is obtained doing

y = 1000 m in y = y_{i} + v_{i}t + ½ at^{2}:

1000 m = 12 m/s·t + ½ ·9,8m/s^{2}·t^{2} or

0 = 4,9t^{2} + 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 v_{i} will be -12m/s and the same acceleration. Or, we
can choose an upward direction axis with v_{i} = + 12m/s and a = -9.8
m/s^{2}.

The usual choice is this last one, i.e.

v_{i}=+12 m/s, a=-9.8 m/s^{2}, y_{i} = 0, t = 10 s

v = 12 m/s - 9.8 m/s^{2}·10s = -86 m/s

y = 12 m/s·10s - ½ ·9.8m/s^{2}·(10s)^{2} = -370 m or 630 m
from ground.

The time to hit the ground is obtained doing

y = -1000 m in y = y_{i} + v_{i}t + ½ at^{2}:

(*) -1000 m = 12 m/s·t - ½ ·9,8m/s^{2}·t^{2} or

0 = 4,9t^{2} - 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/s^{2}·t^{2} , which is the same
equation (*) multiplied by -1 or 0 = 4,9t^{2} - 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**

**Other Related Sites:
· Physics,
Main Page
· Physics,
Mathematics
· Physics,
Detailed Homework Scope Help
· Energy, Work
and Power: Concepts
· Kinetic
Energy
· Potential
Energy
· Power
· Physics
Problems, Example
· Physics
Homework - Mechanical Energy Conservation Problems
· Physics
Homework - Mechanical Power Problems
· Coulomb's
Law
· Exercises
Using Coulomb's Law
· Electric
Field Charges
· Electric
Field Exercises
· Electric
Potential Energy
· Exercises,
Electric Potential Energy
· Ohm's Law,
Principle
· Ohm's
Law Exercises
· Gauss'
Law
· Gauss'
Law Exercises
· Second
Newton's Law
· Second
Newton's Law Examples, Part One
· Second
Newton's Law Examples, Part Two
· Sound
Waves
· Sound
Waves: Standing, Interference, Doppler Effect - Examples
· Sound
Waves, Doppler Effect - Examples
· Vectors,
Scalars
· Vectors,
Scalars - Analytic Method
· Addition
Vector Tools, Problems **