## Calculus – deriving the equations for simple projectile motion

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.)

I simply handed us four equations and used them in a firstsecond… and third post. I said I would show how to derive them.

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.

## First Derivation

Acceleration is the derivative of velocity…

$\frac{d}{dt}v(t) = a(t)$

and that means velocity is the integral of acceleration. We want the acceleration to be constant, so we write a instead of a(t):

$v(t) = \int a \, dt$

and do the integral, supplying a constant of integration when the integral sign disappears:

$v(t) = a\ t + c$.

It is convenient to evaluate the constant at time t = 0, because the term involving t drops out:

$v(0) = a*0 + c$

$v(0) = c$

and so

$v(t) = a\ t + v(0)$.

I learned that in reverse order, and – if you learned this equation – you probably did, too:

$v(t) = v(0) + a\ t$

“V equals v-zero plus a t.”

(I don’t see equations, I hear them.)

Next, velocity is the derivative of position…

$\frac{d}{dt}s(t) = v(t)$

so position is the integral of velocity. That is,

$s(t) = \int v(t) \, dt$

We plug in our formula for v(t)…

$s(t) = \int (a\ t + v(0)) \, dt$

We do the integral, getting another constant of integration…

$s(t) = a\ t^2/2 + v(0)\ t + c$.

Again, we evaluate that formula at t = 0 because two terms will drop out…

$s(0) = a*0 + v(0)*0 + c$

so

$s(0) = c$,

and we have gotten

$s(t) = a\ t^2/2 + v(0)\ t + s(0)$

which I, at least, learned the other way:

$s(t) = s(0) + v(0)\ t + a\ t^2/2$

“S equals s-zero plus v-zero t plus a t-squared over 2.”

## Second Derivation

There is yet another way to do this; use the fundamental theorem of integral calculus, which says:

if

$\frac{d}{dt}F(t) = f(t)$

then

$F(b) - F(a) = \int_a^b f(t) \, dt$.

(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.

$F(t) - F(0) = \int_0^t f(\tau) \, d\tau$.

(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.)

Then from

$\frac{d}{dt}v(t) = a(t)$

with F = v and f = a, we get

$v(t) - v(0) = \int_0^t a(\tau) \, d\tau$.

For constant acceleration, $a(\tau) = a\$, and we get…

$v(t) - v(0) = \int_0^t a \, d\tau$

which is…

$v(t) - v(0) = a\ \tau\ |_0^t$

$v(t) - v(0) = a\ t - a*0 = a\ t$

so finally

$v(t) = a\ t + v(0)$,

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.

Similarly, from

$\frac{d}{dt}s(t) = v(t)$

with F = s and f = v, we write the definite integral…

$s(t) - s(0) = \int_0^t v(\tau) \, d\tau$

Drop in our formula for v(t) – excuse me, $v(\tau)\$

$s(t) - s(0) = \int_0^t (a\ \tau + v(0)) \, d\tau$.

Do the integrals…

$s(t) - s(0) = a\ \tau^2/2 |_0^t + v(0)\ \tau |_0^t$

getting

$s(t) - s(0) = a\ t^2/2 + v(0)\ t$

and we may rearrange that, of course:

$s(t) = s(0) + v(0)\ t + a\ t^2/2$.

## 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.

$s(t) = s(0) + v(0)\ t + a\ t^2/2$

$v(t) = v(0) + a\ t$;

With a = 0, the equations for the horizontal position and speed simplify to

$s(t) = s(0) + v(0)\ t$

$v(t) = v(0)$.

With a = -g, we get the equations for vertical position and speed:

$s(t) = s(0) + v(0)\ t - g\ t^2/2$

$v(t) = v(0) - g\ t$

And that’s all, folks.

### 5 Responses to “Calculus – deriving the equations for simple projectile motion”

1. Jason Says:

So im a bit late to you post, (a year and then some), but I came across this when preping for my final exam. I just wanted to drop a line and say you are very skilled in explaining these concepts in a systematic and clear way. It was a nice read and refresher. Bravo!

2. Bryan Says:

Agreed. Great post!

3. […] An explanation […]

4. Watermelon Says:

Neat explanation. Though, it would’ve been better if you used different notations for the vertical and horizontal velocity – distance components.