Finding an expression for max distance in terms of theta

[Dizzy]

Homework Statement

Basic projectile motion if you throw from ground level, to obtain a max distance in the x you'd want to throw at a 45° angle right? So as your initial height increases, your throwing angle will decrease and my guess is as your initial velocity increases your angle increases but not greater than 90°?? Anyways I want to find this expression of max distance in terms of theta.

Homework Equations

y=y_o+v_osinθ⋅t-4.9t²
x=v_ocosθ⋅t
t=(-b+/-√(b²-4ac))/2a

The Attempt at a Solution

So my attempt at finding this solution is finding t when height is 0 in terms of theta. In this problem I am assume v_o and y_o are constants that I can change when ever I like (seen later in desmos) So I used the quadratic formula to solve for t in terms of theta and those constants. I got t=((-v_osinθ +/- √(v_o²sin²θ+19.6y_o))/-9.8) from now on I will refer to t as the - side of the +/- so t is that whole quadratic formula but not the plus, the minus end because I want the most negative on the top since there is a negative on the bottom to give me a larger positive time. I plugged that into my x=v_ocosθ⋅t. Then I took the derivative of that with respect to theta. In my mind if I find a zero on this derivative function I should be given a critical value which will represent maximum and minimum distances in the x direction. I am going to link my graph in desmos that will show what my derivative came out to be. So this graph is set to be rate of distance in the y-axis and theta as my x-axis. I am getting an angle of π in order to get a max distance when y_o =50m and v_o=4m/s. I want to throw backwards when I am at these initial conditions? This is definitely wrong. Can you help me find my error? Thank you.

EDIT: here is the graph. Sorry I didn't see the edit button you can ignore my reply https://www.desmos.com/calculator/usgxxh13qe

 

[Dizzy]

Homework Statement

Basic projectile motion if you throw from ground level, to obtain a max distance in the x you'd want to throw at a 45° angle right? So as your initial height increases, your throwing angle will decrease and my guess is as your initial velocity increases your angle increases but not greater than 90°?? Anyways I want to find this expression of max distance in terms of theta.

Homework Equations

y=y_o+v_osinθ⋅t-4.9t²
x=v_ocosθ⋅t
t=(-b+/-√(b²-4ac))/2a

The Attempt at a Solution

So my attempt at finding this solution is finding t when height is 0 in terms of theta. In this problem I am assume v_o and y_o are constants that I can change when ever I like (seen later in desmos) So I used the quadratic formula to solve for t in terms of theta and those constants. I got t=((-v_osinθ +/- √(v_o²sin²θ+19.6y_o))/-9.8) from now on I will refer to t as the - side of the +/- so t is that whole quadratic formula but not the plus, the minus end because I want the most negative on the top since there is a negative on the bottom to give me a larger positive time. I plugged that into my x=v_ocosθ⋅t. Then I took the derivative of that with respect to theta. In my mind if I find a zero on this derivative function I should be given a critical value which will represent maximum and minimum distances in the x direction. I am going to link my graph in desmos that will show what my derivative came out to be. So this graph is set to be rate of distance in the y-axis and theta as my x-axis. I am getting an angle of π in order to get a max distance when y_o =50m and v_o=4m/s. I want to throw backwards when I am at these initial conditions? This is definitely wrong. Can you help me find my error? Thank you.

EDIT: whoops I completely forgot the graph here you go https://www.desmos.com/calculator/usgxxh13qe

 

[Ray Vickson]

Homework Statement

Basic projectile motion if you throw from ground level, to obtain a max distance in the x you'd want to throw at a 45° angle right? So as your initial height increases, your throwing angle will decrease and my guess is as your initial velocity increases your angle increases but not greater than 90°?? Anyways I want to find this expression of max distance in terms of theta.

Homework Equations

y=y_o+v_osinθ⋅t-4.9t²
x=v_ocosθ⋅t
t=(-b+/-√(b²-4ac))/2a

The Attempt at a Solution

So my attempt at finding this solution is finding t when height is 0 in terms of theta. In this problem I am assume v_o and y_o are constants that I can change when ever I like (seen later in desmos) So I used the quadratic formula to solve for t in terms of theta and those constants. I got t=((-v_osinθ +/- √(v_o²sin²θ+19.6y_o))/-9.8) from now on I will refer to t as the - side of the +/- so t is that whole quadratic formula but not the plus, the minus end because I want the most negative on the top since there is a negative on the bottom to give me a larger positive time. I plugged that into my x=v_ocosθ⋅t. Then I took the derivative of that with respect to theta. In my mind if I find a zero on this derivative function I should be given a critical value which will represent maximum and minimum distances in the x direction. I am going to link my graph in desmos that will show what my derivative came out to be. So this graph is set to be rate of distance in the y-axis and theta as my x-axis. I am getting an angle of π in order to get a max distance when y_o =50m and v_o=4m/s. I want to throw backwards when I am at these initial conditions? This is definitely wrong. Can you help me find my error? Thank you.

EDIT: here is the graph. Sorry I didn't see the edit button you can ignore my reply https://www.desmos.com/calculator/usgxxh13qe

You need to decide whether to choose the + or - root of your quadratic. You can decide that by noting that the range-time must be greater than the maximum-altitude time (since the projectile needs to fall back from the maximum altitude while still moving forward. Only one of your two roots satisfies that condition. (You might think about what the physical meaning, if any, corresponds to the other root.)

 

[Dizzy]

You need to decide whether to choose the + or - root of your quadratic. You can decide that by noting that the range-time must be greater than the maximum-altitude time (since the projectile needs to fall back from the maximum altitude while still moving forward. Only one of your two roots satisfies that condition. (You might think about what the physical meaning, if any, corresponds to the other root.)

Hi!
Thanks for your reply but I mentioned in my problem that I choose the - root of the quadratic because it will give me the largest negative number over a -9.8 giving me the largest positive time.
Thanks again,
Matt

 

[Ray Vickson]

Hi!
Thanks for your reply but I mentioned in my problem that I choose the - root of the quadratic because it will give me the largest negative number over a -9.8 giving me the largest positive time.
Thanks again,
Matt

You say you are getting an angle of $\pi$ at the maximum distance. Well, just setting the derivative to zero can give either a max or a min; you need to either check additional conditions, or else you need to restrict the angle right from the start. Choosing $\theta \in [0,\pi/2]$ makes sense because that corresponds to throwing the thing up and forward (with $\theta = 0$ meaning you throw it forward and not up at all). If you want to allow throwing it down and forward as well you can restrict your analysis to the interval $\theta \in [-\pi,\pi]$, but obviously, throwing it down must result in a shorter range and so cannot possibly be optimal.

Anyway, when I use your parameters $v_0=4, y_0=50, g = 9.81$ I get an optimal angle slightly positive---throw it up at a surprisingly shallow angle, nowhere near $45^{\circ}$.

For what it's worth: the optimal angle can be obtained explicitly in closed-form, by solving the equation $d\;\text{Range} / d\theta = 0$. However, it is messy and would be time-consuming if done by hand; I used the computer algebra package Maple to do the heavy lifting (plus the numerics and the graphs as well).

 

[Dizzy]

You say you are getting an angle of $\pi$ at the maximum distance. Well, just setting the derivative to zero can give either a max or a min; you need to either check additional conditions, or else you need to restrict the angle right from the start. Choosing $\theta \in [0,\pi/2]$ makes sense because that corresponds to throwing the thing up and forward (with $\theta = 0$ meaning you throw it forward and not up at all). If you want to allow throwing it down and forward as well you can restrict your analysis to the interval $\theta \in [-\pi,\pi]$, but obviously, throwing it down must result in a shorter range and so cannot possibly be optimal.

Anyway, when I use your parameters $v_0=4, y_0=50, g = 9.81$ I get an optimal angle slightly positive---throw it up at a surprisingly shallow angle, nowhere near $45^{\circ}$.

For what it's worth: the optimal angle can be obtained explicitly in closed-form, by solving the equation $d\;\text{Range} / d\theta = 0$. However, it is messy and would be time-consuming if done by hand; I used the computer algebra package Maple to do the heavy lifting (plus the numerics and the graphs as well).

Yes I have solved this problem multiple times and I have even graphed it correctly, I got an angle of about 7° here is the the visual. https://www.desmos.com/calculator/i90cbawcqs ... I understand this is a time consuming problem if I have to do all this work that is why I want to get an equation that could give me theta if I just plugged in what the initial velocity and starting height are. I think the problem with the graph I posted is I did the derivative by hand which gives me some room for error. I am not so worried about the angle π I just want to know if there is anyway I could get an expression for θ in terms of initial velocity and starting height. I am going to keep trying new things out but I just want to know how I should approach making a equation for θ in terms of v_o and y_o.
Thanks for your responses and interest in my problem!

 

[Ray Vickson]

Yes I have solved this problem multiple times and I have even graphed it correctly, I got an angle of about 7° here is the the visual. https://www.desmos.com/calculator/i90cbawcqs ... I understand this is a time consuming problem if I have to do all this work that is why I want to get an equation that could give me theta if I just plugged in what the initial velocity and starting height are. I think the problem with the graph I posted is I did the derivative by hand which gives me some room for error. I am not so worried about the angle π I just want to know if there is anyway I could get an expression for θ in terms of initial velocity and starting height. I am going to keep trying new things out but I just want to know how I should approach making a equation for θ in terms of v_o and y_o.
Thanks for your responses and interest in my problem!

I get the solution $\theta = 0.1260100559 \approx 0.126 \; \text{radians}$, which equals $7.219844378^{\circ} \approx 7.2^{\circ}$, as do you. (I used $g = 9.81$ instead of your $g = 9.8$.)

In general, getting a closed-form expression for the range $X(\theta)$ from initial height $y_0$, initial speed $V$ and initial angle $\theta$ can be done just by substituting the range-time $T$ into the $x$-equation $x = V \cos(\theta) T$. After simplification that gives
$$X(\theta) = \frac{1}{g} V^2 \cos(\theta) \sin(\theta)
+ \frac{1}{g} V \cos(\theta) \sqrt{V^2 \sin^2 (\theta) + 2 g y_0}$$
To get the maximum range you need to find a root of $d X(\theta) / d \theta = 0$. This is messy, but if you are prepared to spend a few hours doing algebra, you can get a solution. Alternatively, you can use a computer algebra system and get a solution in about a millisecond. PF rules will not allow me to write down the final result here, but rest assured, it is do-able, and the final result is not overly complicated.

 

[Dizzy]

I get the solution $\theta = 0.1260100559 \approx 0.126 \; \text{radians}$, which equals $7.219844378^{\circ} \approx 7.2^{\circ}$, as do you. (I used $g = 9.81$ instead of your $g = 9.8$.)

In general, getting a closed-form expression for the range $X(\theta)$ from initial height $y_0$, initial speed $V$ and initial angle $\theta$ can be done just by substituting the range-time $T$ into the $x$-equation $x = V \cos(\theta) T$. After simplification that gives
$$X(\theta) = \frac{1}{g} V^2 \cos(\theta) \sin(\theta)
+ \frac{1}{g} V \cos(\theta) \sqrt{V^2 \sin^2 (\theta) + 2 g y_0}$$
To get the maximum range you need to find a root of $d X(\theta) / d \theta = 0$. This is messy, but if you are prepared to spend a few hours doing algebra, you can get a solution. Alternatively, you can use a computer algebra system and get a solution in about a millisecond. PF rules will not allow me to write down the final result here, but rest assured, it is do-able, and the final result is not overly complicated.

What is against the rules? Writing the actual expression? Also what algebra system did you use to solve this? Everything I tried to use says it does not support my equation.
Thanks!

 

[Ray Vickson]

What is against the rules? Writing the actual expression?

Yes, that is exactly what I said. We are not permitted to supply detailed solutions.

Also what algebra system did you use to solve this?

As I said in #5, I used Maple.

Everything I tried to use says it does not support my equation.

I do not know what this means. How is an equation "not supported"?

Anyway, for given numerical inputs $v_0, g, y_0$ you can find the maximum in EXCEL, using the Solver tool (with $\theta$ in the "variable" cell). Other spreadsheets have similar capabilities.

Thanks!

 

[Dizzy]

Here is what I mean,
When I try to get an equation solver to solve for x, it says that it can not do it. This was the correct formula for our derivative right? So if i set it equal to 0 and assume the rest of the numbers are constants then I should get x in terms of all of those constants but nothing can find that for me.
NOTE: if it is hard to see the image hold ctrl and scroll in, sorry I quickly pasted and cropped it out in MS Paint

 

[Ray Vickson]

Here is what I mean,
When I try to get an equation solver to solve for x, it says that it can not do it. This was the correct formula for our derivative right? So if i set it equal to 0 and assume the rest of the numbers are constants then I should get x in terms of all of those constants but nothing can find that for me.
NOTE: if it is hard to see the image hold ctrl and scroll in, sorry I quickly pasted and cropped it out in MS Paint

I don't know if you have the correct formula for the derivative, since I cannot read your posted image. Just type it out; avoid posting images if at all possible. Sure, it takes a bit of time and is a bit of trouble, but so what? Helpers type out complicated formulas all the time, usually using LaTeX. For example, see post #7. You can also type it in plain text, using u^2 for $u^2$ , sqrt(something) for $\sqrt{\text{something}},$ a/(b+c) for $\frac{a}{b+c}$ etc.