While we are here, it might be fun and instructive to work out the math for the trajectory of a falling object using the calculus tools we have developed.

Experiment has shown that if you drop an object, it starts out at 0 velocity and starts moving downward so that after 1 second it is traveling at 9.8 meters per second (m/s) downward, after 2 seconds it is traveling at 19.6 m/s downward (exactly twice as fast).  And so, the velocity in the downward direction (negative y-direction) is

  v_y(t) = -(9.8 m/s/s) * t

where “t” is the time it has been dropping.

We often use the letter “g” to be the acceleration of gravity g=9.8m/s/s to show how fast things speed up when they are dropped. It is measured in meters per second per second.

Thus, when we drop an object, we then can calculate the speed that it will be falling at every moment “t”:

v_y(t) = -g * t

We want to find the position of how far it has dropped at every moment “t”:

v_y(t) = dy/dt = -g * t
y(t) = ?

We can use differentials and the Integral function from the previous post to figure this out, we drop the object at some height when t=0, “y(0)”, and want to know the height at any time “t”, “y(t)”:

dy = -g * t * dt

I_y(0)^y(t)(dy) = I₀^t(-g*t*dt)

y(t) – y(0) = I₀^t(-g*t*dt)

We can find this Integral if we know the Anti-Derivative of -g*t:

AD(-g*t) = -g*t²/2 + C

Which you can check by taking its derivative again.  And so, we can solve the integral I₀^t(-g*t*dt):

y(t) – y(0) = AD(-g*t)(t) – AD(-g*t)(0)

y(t) – y(0) = (-g*t²/2 + C) – (0 + C)
y(t) – y(0) = -g*t²/2     or
y(t) = y(0) – g*t²/2

which gives you the y(t) coordinate of the object that you dropped from a height y(0) at all values of “t”.  This is proven out by experiment.

Let us figure out Newton’s problem of finding the trajectory (x(t), y(t)) of a cannon ball that is shot in a direction (D_x, D_y) at a speed “S” at the time “t=0”.  We set up the coordinate so that “y” is pointing in the upward direction and “x” is pointing horizontally in the direction the cannon is fired.  We know the velocity at t=0, and the position of the cannon is (0, 0):

(v_x(0), v_y(0)) = (D_x, D_y)*S
(x(0), y(0)) = (0, 0)

So since

(v_x(t), v_y(t)) = (dx/dt, dy/dt)

And the velocity in the x-direction “v_x” does not change v_x(t) = v_x(0) and the velocity in the y-direction is always decreasing v_y(t) = v_y(0)-g*t:

(dx, dy) = (v_x(0)*dt, (v_y(0)-g*t)*dt)

We integrate both sides to get:

(x(t)-x(0), y(t)-y(0)) = (v_x(0)*t, v_y(0)*t–g*t²/2)

(x(t), y(t)) = (x(0)+v_x(0)*t,  y(0)+v_y(0)*t–g*t²/2)

Finally substituting in our starting values from above:

(x(t), y(t)) = (D_x*S*t,  D_y*S*t – g*t²/2)

We have arrived at the trajectory of the cannon ball at every time “t”.  It is a simple formula that depends only on simple math that is easy to calculate.

At this point, you might sense that we can in theory solve for the trajectory of any object if we know the acceleration that is acting on it at any point in space.  This field of study is called kinematics, and it is used to figure out the trajectory of planets in orbit, or trajectories of rocket ships floating through space, or basically the path of any other object in the universe.  This is also what must have been floating through Newton’s head when he realized that these principles were universal.