# Angles/Resultants (maybe?)

1. Apr 10, 2008

### michaeljjt

Hey all,

First off, I am not a student. I am actually a general/homework help tutor. Unfortunately Physics is probably the most difficult subject for me to tutor. That being said, I was attempting to work on this problem with a student of mine, and ran into a great deal of difficulty, mostly because the textbook is 95% theory, and does not give any examples/ situations of what formulas to use; then gives mathematical problems and questions. Since I am not in class, I'm not sure how much the kid I am tutoring is or is not paying attention, because I would have to assume the teacher is going over this material. I actually don't need anyone to answer this for me, but if you could get me started in the right direction, how to approach it, formulas to use, it would be greatly appreciated! I also need to prepare the student for an upcoming test, so the steps are much more important to me than an answer.

1. The problem statement, all variables and given/known data

A cannonball is fired 80 m/s. Find the time airborne, max height and range if it is fired at 30 degrees, 45 degrees, 60 degrees and 90 degrees.

2. Relevant equations

Not particularly sure. Was looking at graphs with resultants. Also possibly d=vt?

3. The attempt at a solution

Had student graph it out with a protractor to get resultant, and find vertical height and horizontal distance with each angle.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 10, 2008

### Dr. Jekyll

The initial velocity can be divided in two components: $$\vec{v_{0}}=\vec{v_{x}}+\vec{v_{y}}.$$

So you get these realtions:

$$v_{x}=v_{0}cos\alpha$$
$$v_{y}=v_{0}sin\alpha.$$

$$v_{y}$$ is a vertical shot, so the time which it needs to reach the ground again will be:

$$g=\frac{\Delta v}{\Delta t} \Rightarrow t=\frac{2v_{y}}{g}=\frac{2v_{0}sin\alpha}{g}.$$

So, this is the time airborne.

Max height can be done using energies:

$$E_{k}=E_{gp} \Rightarrow mv_{y}^{2}=2mgh \Rightarrow h=\frac{v_{0}^{2}sin^{2}\alpha}{2g}.$$

Range is done by using the x component of the initial velocity:

$$D=v_{x} t \Rightarrow D=v_{0}cos\alpha \frac{2v_{0}sin\alpha}{g}=\frac{v_{0}^{2}sin2\alpha}{g}, (sin2\alpha=2sin\alpha cos\alpha).$$

That's all you need.