Using Calculus To Derive The Freefall Formula
The Position Equation (also known as the freefall formula)
S = 16t^{2} + V_{o}t +
S_{o} is often cited in college algebra textbooks. In
this formula, S represents the height (in feet) of an object thrown
into the air at any time t (in seconds) where V_{o} is
the velocity at which the object is thrown upward (or downward) and
S_{o} is the height of the object when it is thrown. Also,
this equation only works with the acceleration of gravity on the
surface of the earth of 32 feet per second each second. The equation
for any gravity and units is S = at^{2} +
V_{o}t + S_{o}.
Where Does This Formula Come From?
This is not some formula a physics or math professor made up out of
the blue, centuries ago. Rather, this is a formula that is a direct
result of integrating the acceleration function representing gravity.
From calculus, we can write acceleration as a = s''(t)
where s''(t) is the second derivative of distance with respect to time
(i.e. the rate of the change in velocity with respect to time) and
a=32 ft/sec2 represents the acceleration due to gravity.
Integrate both sides of s''(t) = 32
with respect to t to get
This results in s'(t) = 32t + C_{1}
where C_{1} is a constant and s'(t) is the velocity
at any time t.
At time t=0, s'(t) is equal to the initial velocity V_{o}
so we get C_{1} = V_{o}. The new equation is
s'(t) = 32t + V_{o}
Now, integrate both sides with respect to t to get
This results in
where C_{2} is a constant and s(t) is the position at any
time t.
At time t=0, s(t) is equal to the initial position S_{o} so
we get C_{2} = S_{o}. The final equation is
or simplified
Does none of this make sense? With successful
completion of Calculus I & II this would make complete sense.
Furthermore, you can not really understand physics unless you fully
understand how the formulas are derived.
Remember: Physics = Mathematics Applied To
The World Around Us
