A cannon has a muzzle speed velocity Vo of 60.0m/s, at what angle theta should it be aimed to strike a distance 320 meters away. Ignore air resistance

So I get how to set this up but I am confused about one thing: shouldn't acceleration due to gravity be negative? However, whenever I plug in -9.8m/s$^2$ I end up with a negative angle which can't be right because the cannon is shooting the ball up into the air.

This is what I have so far ($R$= horizontal range)

$$R=\frac{V_0^2 \sin2\theta}{g}$$

$$\sin~2\theta = \frac{Rg}{V_0^2}$$

$$2\theta=\sin^{-1}\left[\frac{320\text{ m}\times-9.8\text{m/s}^2}{3600(\text{m/s})^2}\right]$$

the only way I can get the right answer is if I set acceleration due to gravity as positive. Can someone explain why it must be positive in this case?

  • 1
    $\begingroup$ Hi and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Sep 25 '15 at 7:19
  • 1
    $\begingroup$ I know that, I'm not asking for a solution or for someone to check my work, i already have the solution. I want to know why acceleration due to gravity is positive in this circumstance. $\endgroup$ – user93763 Sep 25 '15 at 8:12
  • $\begingroup$ OK I guess our usual casual approach to sign conventions can make things less obvious than they need be :-). I'll add an answer explaining what's going on. $\endgroup$ – John Rennie Sep 25 '15 at 8:48

Physicists tend to be a bit casual about sign conventions when it seems to be obvious. So let's attempt to be completely rigourous.


The key step is getting the flight time $t$ since the range is just $v\cos\theta\, t$. We do this using the SUVAT equation:

$$ v = u + at $$

We'll use the usual conventions that up and right are positive, so $v_y$ and $v_x$ are both positive. The gravitational acceleration points downwards, so it has a negative value. However there are two ways we can treat this. Suppose we set $g = -9.81$ m/s$^2$, then our equation becomes:

$$ -v\sin\theta = v\sin\theta + gt \tag{1} $$

and rearranging for the time gives:

$$ t = \frac{-2v\sin\theta}{g} $$

If we put $g = -9.81$ m/s$^2$ we get a positive value for $t$. So far so good.

But we could also take $g$ to be the magnitude of the gravitational acceleration so it is always a positive quantity. In that case we have to write equation (1) as:

$$ -v\sin\theta = v\sin\theta - gt \tag{2} $$

Note that we now have a minus sign on the right hand side. We put this in because we are taking $g$ to be a magnitude. Rearranging this gives:

$$ t = \frac{2v\sin\theta}{g} $$

So we get the same equation for $t$ as before, but without the minus sign.

When you're looking at this sort of system you have to know what sign convention has been used. Has $g$ been taken to be $-9.81$ m/s$^2$ or has it been taken to be $+9.81$ m/s$^2$ and the minus sign inserted in the equation instead?

If we go back to equation (1) and multiply by $v\cos\theta$ then rearrange to get the range we end up with:

$$ d = \frac{-v^2\sin2\theta}{g} $$

and obviously equation (2) gives us the same without the minus sign:

$$ d = \frac{v^2\sin2\theta}{g} $$

And there you have the answer to your question. The range equation you cite has been derived by taking $g$ to be positive and putting the minus sign in the SUVAT equation instead. So you need to put $g$ in as a positive number or you'll get the wrong answer.

  • $\begingroup$ +1 for general goodness, both in the answer and the change of heart:) $\endgroup$ – user81619 Sep 25 '15 at 10:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy