# Projectile Motion of shot cannon ball

[SOLVED] Projectile Motion

## Homework Statement

A cannon shoots a ball at an angle theta above the ground. Use Newton's second law to find the ball's position as a function of time g(t). What is the largest value of theta if g(t) is to increase throughout the ball's flight.

## The Attempt at a Solution

Alright so I've found the distance squared (as advised in the original problem).
$$g(t) = r^2 = 1/4g^2 t^4 - (v_0 g sin \theta ) t^3 + {v_0}^2 t^2$$
Which I have verified is correct.

So then like usual, I differentiate g(t) with respect to time to find the critical points.
$$g' = g^2 t^3 - 3(v_0 g sin \theta t^2 + 2 {v_0}^2 t$$
Then I use the quadratic formula to find t and set that equal to 0

$$\frac{3v_0 g sin \theta +- \sqrt{9 {v_0}^2 g^2 sin^2 \theta - 8g^2 {v_0}^2} }{2g^2}$$

This is where the problem comes in...
If I were to set g' equal to 0 and solve
$$0 = 3v_0 g sin \theta +- \sqrt{9 {v_0}^2 g^2 sin^2 \theta - 8g^2 {v_0}^2}$$
But then if I solve to 0 I get $$8g^2 {v_0}^2 = 0$$

I know the answer involves setting the discriminant equal to 0, but I don't understand why I can't find the critical points.
So basically my question is how to find theta max? Generally I would find the critical points and then determine if it's a max or a min. In this case, it doesn't seem to work.

Last edited:

Sorry I don't really understand the question.
Could you rephrase it maybe?

Yeh, the question is not very clear...when a golf ball is hit, its distance is naturally going to increase from the origin, then where does the maximum angle come in?

Oh sorry guys, I like to look at my latex code as I type the question, it's done now let me know if it's still unclear, I can type the complete question from the book if necessary.

Oh and one more thing to note is that g is gravity and g(t) is the function, sorry didn't mean to confuse anyone with that notation. I would edit my post but my ability to edit things seems to have disappeared

alphysicist
Homework Helper
Hi jesuslovesu,

I think you made a error in using the quadratic formula at one point. After squareing and taking the derivative, your equation was:

So then like usual, I differentiate g(t) with respect to time to find the critical points.
$$g' = g^2 t^3 - 3(v_0 g sin \theta t^2 + 2 {v_0}^2 t$$
You then wanted to set this equal to zero. The first thing was you cancelled out one of the t's; this meant that t=0 is a solution (but not useful in answering the problem).

Next you used the quadratic formula, but you misused the result. In using the quadratic formula, the question the quadratic formula is asking, "What value of t makes g' equal to zero?" And the answer is:

$$t=\frac{3v_0 g sin \theta +- \sqrt{9 {v_0}^2 g^2 sin^2 \theta - 8g^2 {v_0}^2} }{2g^2}$$

So that is the value of t that makes g'=0; you don't want to then set that equal to zero.

Now that you have the value of t that makes g'=0, you can notice that if theta is too small then the part under the square root will be negative--meaning the time t will be partly an imaginary number. So when theta is too small, there is no physical t (except for t=0) that makes g'=0.

That's exactly what you want. For small angles, there is no critical turning point in g' during the trajectory, so the distance is always increasing. By setting the discriminant equal to zero, you can find the smallest angle at which there is a critical point in g' (or in terms of the question wording, find the largest angle at which there is no critical point).