# Varying the height of a projectile and effect on range-experiment

I am investigating the effect of varying the height of which a ball is rolled down a slope on the range of the projectile (see image attached for experiment set up). I am getting really stuck on determining the mathematical relationship between horizontal range of the projectile and the height the slope (assuming conservation of energy).

This is the set up of of my experiment. Am I supposed to assume that the projectile leaves the table at an angle? According to my data, there is a linear relationship between h and x (which I don't think is true, except I can't figure out what it is actually supposed to be).

Can anyone offer some help here?

EDIT: Sorry the picture is slightly wrong ( I took it from the internet). I have done a quick sketch of what it actually looks like. Sorry for the poor quality!

• Is your table horizontal? If so, the the ball is moving horizontally when it leaves the surface of the table. Aug 26, 2021 at 7:00
• Yes the table is horizontal, but I have made some changes to the diagram. Would it still be leaving the table horizontally? Aug 26, 2021 at 10:49
• I am so sorry but Unfortunately I only have one photo of the set up (it has already been deconstructed) and it has a photo of my ID in it, since I needed proof that I was conducting the experiment (for my teacher to make sure we actually did it ourselves). I am aware that this is super inconvenient- my apologies!!! Aug 26, 2021 at 11:03
• Oh. That's very different! Aug 26, 2021 at 11:09

There is a possibility that energy is lost when the ball collides with the bottom of the ramp given the steepness of the incline. Based on the first diagram, there does not appear to be a reason to think that the projectile leaves the table at an angle, but instead moves horizontally.

But, looking at the added (second) diagram, the ball colliding with bottom of the ramp, also appears to be providing some lift to the ball (by bouncing) since it leaves the edge with what appears to be an upward component of velocity. This could be what is complicating your results, so it is perhaps a good idea to minimize the height the ball is released (and if possible, the incline angle) so that the relationship between $$h$$ and $$x$$ is more accurately given by the derived relation below.

First, let us consider the total energy of the system.

When the ball when the ball reaches the bottom of the ramp, the total energy is given by $$\tag1\frac{1}{2}mv^2+\frac{1}{2}I\omega^2=mgh$$ since the object has both translational and rotational kinetic energy.

If we call $$t$$ the time of flight of the projectile (when it leaves the table to the ground), then we have in horizontal direction $$x=vt$$ or $$t=\frac{x}{v}$$ and since $$H=\frac{1}{2}gt^2$$ since $$v$$ is the horizontal velocity in equation (1) then $$H=\frac{gx^2}{2v^2}$$ or $$v^2=\frac{gx^2}{2H}$$

You can substitute this expression into equation (1) and this will give you an expression for the horizontal range $$x$$ in terms of $$v$$. You can also use the moment of inertia for the ball (a sphere), $$I=\frac{2}{5}mR^2$$ and $$\omega=\frac{v}{R}$$ in equation (1).

You should get something similar to $$x= \sqrt{\frac{20}{7}hH}$$ or $$x=c\sqrt{{h}}$$ for some constant $$c$$ (given that $$H$$ is fixed and assuming I have done the algebra correctly) which looks about right when looking at the diagram. This certainly is not a linear relationship between $$x$$ and height. Perhaps what was meant was there is a linear relationship between $$h$$ and $$x^2$$?

• Thanks for the help. The algebra makes sense, unfortunately my data does not for some unknown reason (perhaps due to the small height measurements as suggested by @silverrahul above). Assuming I my data was correct, from your calculations then, should I be getting a trendline on my graph that runs through the origin? Aug 26, 2021 at 10:47

Is this a numerical problem from a physics textbook or are you actually doing the experiment ?
If it is the former, then you can assume the projectile leaves the table perfectly horizontally.
Of course, in the real world, nothing is perfect and there will always be some slight angle.

According to my data, there is a linear relationship between h and x

This is again something that might just be approximately true in the real world, for small values of h and x.
In general, they will not be linearly related. Rather h and x^2 will be linearly related.

• Thanks for the help! I am actually conducting this experiment in real life. I actually am using pretty small height increments (due to equipment limitations), so are you suggesting that this is perhaps the reason for the somewhat linear relationship?? Aug 26, 2021 at 10:42
• This is incorrect. It should be that $h\propto x^2$. Aug 26, 2021 at 17:15
• @physicslover111 Yes, that is why you are getting linear relationship. Aug 26, 2021 at 17:18
• @Sandejo My mistake, you are right. I mixed them up Aug 26, 2021 at 17:20

For realistic case I "made" from the incline curve a slide curve.

where

$$F(x)=\frac {1}{54}\,x^3-\frac{1}{12} x^2-\,x+5\\$$

at $$x_e=6~$$ is $$F(x)=0$$ and the $$\frac{dF}{dx}\bigg|_{x=x_e}=0$$ thus the velocity is horizontal.

with $$T=\frac 12 m\,v^2+ \frac 12\,\frac 25\,m\,r^2\,\dot\varphi^2\\ U=m\,g\,F(x)\\ v=\frac{dF}{dx}\,\dot x$$ and the roll condition $$\dot x=r\dot\varphi$$ you obtain the energy $$~E=T+U=E(\dot x,x)$$

from the conservation of the energy you have $$E(\dot x,x)=E(\dot x=0,x=0)=m\,g\,h$$ from here you obtain $$v_e=\dot x(x)\bigg|_{x=x_e}$$

you know now the velocity at the end of the slide $$~v_e~$$ you can use now the projectile equations

$$X=v_e\,t\\ Y=H-\frac 12\,g\,t^2\\ \Rightarrow\\ Y(X)=H-\frac 12 \frac{g\,X^2}{v_e}$$ $$Y(X)=0~\Rightarrow~X=\sqrt{\frac{2\,H}{g}}\,v_e\\$$

Edit

you can parametrize the function $$F(x)$$

$$F(x)={\frac { \left( -4\,h+4\,e \right) {x}^{3}}{{e}^{3}}}-{\frac { \left( -9\,h+8\,e \right) {x}^{2}}{{e}^{2}}}+{\frac { \left( -6\,h+4\,e \right) x}{e}}+h$$

where $$~h=F(0)~$$ and e is the parameter where $$F(e)=0$$

from the conservation of the energy at point e you obtain

$$\frac 15\,m\,v_e^2=m\,g\,h$$

hence $$v_e=\sqrt{5\,g\,h}$$

$$X=\sqrt{\frac{2\,H}{g}}\,v_e=\sqrt{10\,H\,h}$$

the result from @joseph.h is $$~X=\sqrt{\frac{20}{7}\,H\,h}~$$ which is good approximation.