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.

The horizontal equations, then, start with

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.

7 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. Watermelon Says:

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

  4. Lisa Says:

    So helpful as I’m doing my Math IA (it’s an IB stuff, like an elementary version of paper)

  5. panadu Says:

    indeed these are founds in mathematical mechanics and can be applied important theorems of Mathematical analysis for conclusions observed prof dr mircea orasanu and prof drd horia orasanu and is normal as followed profound aspects that attract results in CONSTRAINTS OPTIMIZATIONS ,CHASLES Geometry GERGONNE Geometry and Consequences we have left the sum for the first derivative unchanged. The point here is that what the differential equation contains is , and the expression for this must be written as a sum in xn. For this reason we shift the dummy variable in the series for the first derivative by 2:


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: