A spring-gun projects a small rock from the ground with speed $v_0$ at an angle $\theta_0$ above the ground. You have been asked to determine $v_0$. From the way the spring-gun is constructed, you know that to a good approximation $v_0$ is independent of the launch angle. You go to a level, open field, select a launch angle, and measure the horizontal distance the rock travels. You use $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ and ignore the small height of the end of the spring-gun's barrel above the ground. Since your measurement includes some uncertainty in values measured for the launch angle and for the horizontal range, you repeat the measurement for several launch angles and obtain the results given in [the attached plot]. Ignore air resistance.

enter image description here

(a) Select a way to represent the data well as a straight line.

(b) Use the slope of the best straight-line fit to your data from part (a) to calculate $v_0$.

(c) When the launch angle is $36.9^\circ$, what maximum height above the ground does the rock reach?

Part (a): I've computed as a least-squares linear fit for the data

$$x(\theta)=9.34\,\mathrm m+\left(0.002\,\frac{\mathrm m}{\mathrm{deg}}\right)\theta$$

but on its own, this doesn't seem very useful because (1) the relationship between the distance $x$ and the launch angle $\theta$ clearly doesn't look linear (in fact we know that $x=\frac{{v_0}^2\sin2\theta}g$), and (2) I don't see how I could possibly get information about $v_0$ from this model.

enter image description here

So what is the linear relationship I should be looking for?

Edit to Part (a)

Taking the suggestion from the comment/answer, I've plotted $x$ with respect to $s=\sin2\theta$:

enter image description here

where the equation of the line is

$$x(s)=0.147\,\mathrm m+(10.9\,\mathrm m)s$$

Part (b):

The coefficient of $s$ above gives the line's slope, and comparing this to the coefficient we expect for $\sin2\theta$, I solve for $v_0$:

$$\frac{{v_0}^2}g=10.9\,\mathrm m\implies v_0=10.3\,\frac{\mathrm m}{\mathrm s}$$

Part (c): Using the answer from part (b), I find the maximum height via

$$-(v_0\sin36.9^\circ)^2=-2gy_{\rm max}\implies y_{\rm max}=1.96\,\mathrm m$$

To summarize, my questions are

  • What's going on in part (a)? That is, what kind of answer is most likely expected? (Alas, my textbook only offers solutions for even-numbered exercises.)

  • Are the solutions for parts (b) and (c) reasonable?

  • $\begingroup$ But the part a is already solved, you have to demonstrate that there is a linear relation between $x$ and $\sin(2\theta)$ (that formula is written in the beginning) and plot $x$ vs $s=\sin(2\theta)$... $\endgroup$
    – user153036
    Oct 16, 2018 at 21:34
  • $\begingroup$ @santimirandarp I see, I assumed (given the plot of the data) that I was supposed to find a linear relationship between $x$ and $\theta$, not $\sin2\theta$. $\endgroup$
    – user170231
    Oct 17, 2018 at 2:15
  • $\begingroup$ no problem, difficult and interesting exercise :) $\endgroup$
    – user153036
    Oct 17, 2018 at 2:21
  • $\begingroup$ @user170231, did your teacher show you how to linearize an equation? THAT is what part (a) is all about. $\endgroup$ Oct 17, 2018 at 15:52
  • $\begingroup$ No teacher involved here, just me with a textbook, though I do have a background in math. My fault in part (a) has been addressed, I think. I had assumed the linear relationship was to be established directly between $x$ and $\theta$, and had not considered one between $x$ and $\sin2\theta$. $\endgroup$
    – user170231
    Oct 17, 2018 at 15:59

1 Answer 1


Try defining a new parameter, S, where S = sin(2θ). Now the data looks like a straight line function of S.

  • $\begingroup$ But isn't $s=\sin2\theta$ dimensionless? What would be the physical interpretation of the slope of this line or $\frac{\mathrm dx}{\mathrm ds}$? $\endgroup$
    – user170231
    Oct 16, 2018 at 20:45
  • $\begingroup$ Your own equation for x in terms of Vo, g, and sin(2x) should tell you. $\endgroup$
    – S. McGrew
    Oct 16, 2018 at 23:56
  • $\begingroup$ Sorry, I don't want to give you the answer, Just want to give you hints. $\endgroup$
    – S. McGrew
    Oct 16, 2018 at 23:57
  • $\begingroup$ Thanks, the hint is appreciated, though I'm still somewhat unsure about how to reconcile the units of the slope of the line - like I said, I would expect $s$ to have no units associated with it, but maybe I'm wrong - with the velocity in m/s... Is it as simple as taking the derivative of $x(s)$ with respect to time? $\endgroup$
    – user170231
    Oct 17, 2018 at 2:27
  • 1
    $\begingroup$ The slope of a line is "the rise over the run". In the data graph you showed, the slope would be in meters per degree. If you graph x vs sin(2 theta), the slope will simply be in meters. A best fit to your data will give you a value for Vo^2/g (in meters) You know the value of g, so can easily solve for Vo (units of meters/sec). You need to think some more about how to solve part c. What formula gives the height reached, given a particular Vo and theta? $\endgroup$
    – S. McGrew
    Oct 17, 2018 at 4:59

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