# Energy and Projectiles?

I post here very infrequently, but I'm a high school physics teacher occasionally encountering excellent student questions. Here is one I received today:

Since we've been using energy conservation to simplify problems that were annoying earlier in the year (kinematics, mechanics, etc.), can we apply this to projectiles? (Me: yes, definitely, here's some examples...) Particularly, deriving the range formula is annoying, can we do this with energy? (Me: can't think of how...) Also, finding the range of a projectile shot at an angle above the horizontal off a cliff of a known height is really annoying, how about using energy for this? (Me: can't think of how...)

Any thoughts, folks?

## Answers and Replies

Simon Bridge
Science Advisor
Homework Helper
Its worse with energy... think: principle of least action.
Basically kinematics is good for geometry while physical laws are good for physics.

I think you need to start by defining the problem: what is it about deriving the range formula which is considered annoying?
Why are students deriving this formula anyway, if they already have kinematic equations and/or can draw velocity time graphs?
Once you can articulate what the exercize is for and how students relate to it, you can make progress.

You may try getting range from the position that initial kinetic energy is restored, but is that an improvement?

Khashishi
Science Advisor
It's easiest to directly use the forces in this case. Calculating from the energy can be done but it is not a shortcut. If you use the Hamiltonian formulation of classical mechanics, http://en.wikipedia.org/wiki/Hamiltonian_mechanics
For a simple projectile falling in gravity problem, the Hamiltonian is the energy.
When you calculate the change in canonical momentum (equal to the momentum in this case), you basically end up deriving the Newtonian forces anyways.
So, it's pointless in this case.

Simon Bridge
Science Advisor
Homework Helper
There's a less formal approach ...
I was thinking: take the projectile with initial speed ##u## at angle ##\theta## to the horizontal.
Taking ##x## for horizontal and ##y## for vertical, per usual: the velocity is ##\vec v(t)= \hat\imath u\cos\theta + \hat\jmath (u\sin\theta - gt)##

From there you'd get the max height ##h## from finding ##h: mgh=\frac{1}{2}mu_y^2## right?

To get the range requires some extra prep.
Notice that ##x(t)=ut\cos\theta##, (from d=vt) so I can rewrite the velocity in terms of the horizontal position: $$\vec v(x) = \hat\imath u\cos\theta + \hat\jmath \left(u\sin\theta - \frac{gx}{u\cos\theta}\right)$$ ... a secondary student may prefer vectors written as components on separate lines to the unit vectors above.

Thus: kinetic energy varies with horizontal position as: $$K(x)=\frac{1}{2}m\left[u^2\cos^2\theta + \left(u\sin\theta -\frac{gx}{u\cos\theta}\right)^2\right]$$
Taking the range ##R## to be the horizontal distance where the projectile arrives at it's initial height (i.e. starts on the ground, lands on the ground), then put ##K(R)=K(0)## and solve for ##R##.

Like I said: not really an improvement - unless the person who wants it finds it easier to think like this than draw graphs.
To be fair, it reads better if you separate out the vertical and horizontal contributions to the total KE.