We have worked several examples of simple projectile motion – meaning that the acceleration of gravity was constant and vertical, and there was no acceleration in the horizontal direction. (In particular, there is no air resistance.)
It was junior year in high school that I learned the equations for position and speed as a function of a constant acceleration. I didn’t take calculus until I was a college freshman… and at some point I decided that I knew enough calculus to derive the equations I had been told to memorize two years before.
This is about as elementary as it gets in calculus, but when it was all new to me, it was a thrill to see what it could do for me in physics. I will actually derive them in two slightly different ways.
Here we go.
Acceleration is the derivative of velocity…
and that means velocity is the integral of acceleration. We want the acceleration to be constant, so we write a instead of a(t):
and do the integral, supplying a constant of integration when the integral sign disappears:
It is convenient to evaluate the constant at time t = 0, because the term involving t drops out:
I learned that in reverse order, and – if you learned this equation – you probably did, too:
“V equals v-zero plus a t.”
(I don’t see equations, I hear them.)
Next, velocity is the derivative of position…
so position is the integral of velocity. That is,
We plug in our formula for v(t)…
We do the integral, getting another constant of integration…
Again, we evaluate that formula at t = 0 because two terms will drop out…
and we have gotten
which I, at least, learned the other way:
“S equals s-zero plus v-zero t plus a t-squared over 2.”
There is yet another way to do this; use the fundamental theorem of integral calculus, which says:
(All that says is that the area under the curve f(t), between the points t = a and t = b, is given by the antiderivative F(t) evaluated at the points a and b. It relates the integral of f to the area under the curve f.)
In particular, we may take a = 0 and b = t.
(Yes, I have changed the variable of integration. I’m using “t” for the upper limit of a definite integral, and I cannot simultaneously use the symbol dt.)
with F = v and f = a, we get
For constant acceleration, , and we get…
just as before. This time we didn’t need to evaluate an unknown constant of integration; v(0) was there from the beginning, on the LHS.
with F = s and f = v, we write the definite integral…
Drop in our formula for v(t) – excuse me, …
Do the integrals…
and we may rearrange that, of course:
Specializing to simple projectile motion
Finally, for simple projectile motion, we take the horizontal component of acceleration to be zero, and the vertical component of acceleration to be -g.
The horizontal equations, then, start with
With a = 0, the equations for the horizontal position and speed simplify to
With a = -g, we get the equations for vertical position and speed:
And that’s all, folks.