Kinematics:Free-fall
Naturally, objects fall. This is due to gravity of the earth. In all what it
will follow, we assume that air resistence has a negligible effect on a falling
object, so that the approximation of the object's acceleration is due
entirely to gravity is valid. Inthis case the fall is called free-fall motion.
Galileo Galilei (1564-1642) made quantitative studies to find that the acceleration
due to gravity is constant during the fall. Further, this acceleration is the same
for any falling object. The magnitude of this accelation is represented by the
symbol g and has the value of 9.8 m/s2.
1. Free-fall acceleration
To describe free-fall, we choose the y-axis along the direction
of motion with a vector unit j directed upward. Therefore, the
acceleration of a free-fall object is a = - g j.
free fall vector acceleration:
The sign minus is explicit. It choose only that
the acceleration "a" is downward whereas the unit vector j is upward.
the value g = 9.8 m/s2 is positive.
2. Equations of free-fall
2.1. General formula
Since the acceleration a is constant, we can write the
following vector equations :
a = dv/dt
Integrating with respect to t to gives:
v(t) = at + vo.
(1)
A second integral gives:
y(t) = (1/2) a t2 + vot + yo (2)
If we choose + y upward, then a = - g. yo and
vo depend on their initial values and
intial directions .
Therefore
the general eqation for a free-fall motion is:
y(t) = - (1/2) g t2 + vot + yo
general free fall motion equation:
y(t) = - (1/2) g t2 + vot + yo
Inserting t given by the equation (1) into the equation (2) gives
v2 - vo2 = -2g(y - yo)
To recap:
Related free fall motion equations:
v(t) = - gt + vo
y(t) = - (1/2)gt2 + vot + yo
v2 - vo2 = - 2g(y - yo)
2.2. Equation of free-fall
Without initial velocity, vo = 0, and by choosing the
origin of distances at the point yo = 0, we obtain:
y(t) = - (1/2) g t2
free fall equation:
object released from rest at y = 0
y(t) = - (1/2) g t2
2.3. Examples:
Example 1
If an object is thrown vertically upward with initial
speed vo. We want to determine (a) the maximum height
the object can reach and (b) the time required for the object to
reach this maximum height.
Use the following formula:
v(t) = -g t + vo
v2 - vo2 = - 2g(y - yo)
Let the origin of the coordinate frame be at the release point, and
t = 0 the instant the ball is released. So o = 0. At the time
tmax the object reach it maximum height, its speed is zero.
Then the latter two equations become:
0 = - g tmax + vo
- vo2 = - 2gymax
Therefore:
tmax = vo/g, and
ymax = vo2/2g
Example 2
A stone is thrown upward at a velocity of speed vo = 25 m/s, from the base of a cliff 25 m high. How high is the
highest point of the stone from the top of the cliff?
We use the formula:
v(B)2 - v(A)2 = 2 a (x(A) - x(B))
from the highest point B to the ground A. That is:
02 - vo2 = - 2 g ( 25 + h - 0)
or:
- vo2 = - 2 g (25 + h)
Therefore:
h = (vo2 /2g) - 25 = (25)2 /2 x 9.8) - 25 = 6.9 m.
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