0
$\begingroup$

I used this website to help me generate displacement parametric equations for projectile motion. I have these equations:

$x= ut\cos(θ)$ and $y=ut\sin(θ)-\frac{1}{2}gt^{2}$

Where $x$ is the distance travelled from the origin, $y$ is the vertical position of the projectile, $u$ is the initial velocity, $t$ is the time elapsed and $θ$ is the angle of launch from the origin.

An AK-47 has a intitial projectile velocity of $u=715ms^{-1}$ and a launch angle of $θ=45°$.

I have assumed for this equation $g=9.81ms^{-2}$.

My question is do I need to square the $g$ value? The reason I ask this is as gravity is measured in $ms^{-2}$ and the velocity is measured in $ms^{-1}$.

If I leave the value for g in this equation unchanged as $g=9.81$ then it takes the projectile approximately $103.07s$ to land back on the ground, with a maximum distance of approximately $52.133km$. This seems unrealistic, given how in many games and videos the bullets appear to drop very quickly: Graph where g is not squared.

If I square the value of $g$ then it takes the projectile approximately $10.51s$ to land back on the ground, with a maximum distance of approximately $5.312km$. This seems to be a much more realistic figure and seems to give a much more accurate curve. Without squaring it, it appears as if gravity has no effect on the projectile (which appears to not be the case in many games and many videos.) Graph where g is squared.

According to this wikipedia page The AK-47 has an effecitve range of $350m$, a horizontal range of $530m$, a lethal range of $1500m$ and a maximum range of $2300m$. You can also see the curve of a projectile on this graph: enter image description here

Hence, I am led to believe that $g$ is squared, and I would like confirmation of such. Thank you. I am aware this question may seem a little silly, but I couldn't find anything specifically that stated whether or not g is squared or not.

$\endgroup$
5
  • $\begingroup$ Gravity is the weakest of the fundamental forces and by many orders of magnitude. It is in fact extremely weak. $\endgroup$ Apr 1, 2021 at 20:11
  • $\begingroup$ @CyclotomicField: That is completely irrelevant in this context. Nobody is asking how the strong nuclear force influences the trajectory of a macroscopic bullet. $\endgroup$
    – TonyK
    Apr 1, 2021 at 20:12
  • $\begingroup$ @TonyK he said he found it counter-intuitive because gravity isn't that strong. He should expect to be weak and the math should support that. $\endgroup$ Apr 1, 2021 at 20:14
  • $\begingroup$ But @CyclotomicField, gravity and air resistance are the only forces that come into play here! Are you just trying to confuse the OP? $\endgroup$
    – TonyK
    Apr 1, 2021 at 20:17
  • $\begingroup$ Also note that very few people fire a gun at 45 degrees over long range, and that 45 is the optimum angle for distance of a projectile $\endgroup$
    – Henry Lee
    Apr 1, 2021 at 20:41

2 Answers 2

1
$\begingroup$

Okay so your equations are correct, if we neglect air resistance, spin etc. It comes from: $$x''(t)=0$$ $$x'(t)=C_1$$ $$x(t)=C_1t+C_2$$ This is due to the fact that there are no forces acting in the $x$ direction and so using $F=ma$ we get $0$ acceleration. Now the velocity is constant throughout, which will be equal to the initial velocity of $v_0\cos(\theta_0)$. The displacement has a constant which will be the initial displacement in $x$, which we can call $x_0$ and so: $$x(t)=v_0\cos(\theta_0)t+x_0$$


For the $y$ we know that the only force acting is that due to gravity, $F=-mg$ and so $-mg=ma$ giving $a=-g$, giving: $$y''(t)=-g$$ $$y'(t)=-gt+C_3$$ $$y(t)=-\frac{g}{2}t^2+C_3t+C_4$$ now we know that $C_3$ is just the initial velocity in $y$ and $C_4$ is initial displacement in $y$ and so: $$y(t)=-\frac{g}{2}t^2+v_0\sin(\theta_0)+y_0$$


We now have you system of equations: $$\begin{cases}x=v_0\cos(\theta_0)t+x_0\\y=-\frac{g}{2}t^2+v_0\sin(\theta_0)+y_0\end{cases}\,\,\,t\ge0$$ You can actually rearrange $x$ for $t$ and substitute into $y$ to get a function $Y(x)$ which shows the motion is parabolic. note that in your case $x_0=y_0=0$


This shows why the quantity is $g$ and not a power of $g$. In terms of the units, It has the unit of acceleration, the second derivative of displacement with respect to time and hence takes the form $ms^{-2}$.

As I said at the start, these equations are entirely based on air resistance being negligible, which for fast moving objects is far from true. In fact, a common model for air resistance (drag) is: $$F_D=\frac12\rho C_DAv^2$$ Notice how drag is considered roughly proportional to the square of velocity. This means at the point of highest velocity (when it leaves the gun) the deceleration is at its greatest and then tapers off as velocity drops. If you wish to use this model the equations would be: $$m\frac{d^2}{dt^2}\begin{pmatrix}x\\y\end{pmatrix}=-g\begin{pmatrix}0\\1\end{pmatrix}-\frac{\rho C_dA\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}}{2}\frac{d}{dx}\begin{pmatrix}x\\y\end{pmatrix}$$ where: $$x(0)=x_0,y(0)=y_0,x'(0)=v_0\cos(\theta_0),y'(0)=v_0\sin(\theta_0)$$ and in-fact I wrote an entire article on this considering variable gravitational force and density/viscosity of air.

$\endgroup$
2
  • $\begingroup$ Ah, the $F_{D}=\frac{1}{2}pC_{D}Av^{2}$ bit really confuses me, when putting it in terms of x and y parametric equations, I think I spent an hour just trying to decipher the wikipedia page on this as I found it really hard to understand and physics isn't my strong point since I didn't get the chaneg to study it. Do you mind explaining further (or giving the link to the article?) in which case I can get back to you if I don't understand ... :) $\endgroup$
    – tukars
    Apr 1, 2021 at 22:12
  • 1
    $\begingroup$ @tukars so all of the values can be considered constants apart from $v^2$. Now if we want to calculate this we need to know the magnitude of $v$ since the velocity in $x$ is simply $\frac{dx}{dt}$ and the velocity in $y$ is $\frac{dy}{dt}$ we can now use pythagoras' theorem to get this from vector form to a magnitude, giving $|v|=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}$ $\endgroup$
    – Henry Lee
    Apr 6, 2021 at 12:37
1
$\begingroup$

No, $g^2$ is never something that needs to be used. The discrepancy is entirely due to air resistance. If there was no atmosphere, I could easily believe that an AK-47 would have a range of more than 50km.

$\endgroup$
2
  • $\begingroup$ How would I model air resistance? I would like to make a semi-realistic solution to this problem $\endgroup$
    – tukars
    Apr 1, 2021 at 20:22
  • $\begingroup$ @tukars I have written a thorough answer on the model $\endgroup$
    – Henry Lee
    Apr 1, 2021 at 20:32

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.