Projectiles launched at complementary angles

[Bunny-chan]

Homework Statement

a) Show that for a given velocity $V_0$ a projectile can reach the same range $R$ from two different angles $\theta = 45 + \delta$ and $\theta = 45 - \delta$, as long as $R$ doesn't go over the maximum range $R_{max} = \frac{V_0^2}{g}$. Calculate $\delta$ in function of $V_0$ and $R$.

b) Generalize your previous result, showing that a projectile launched from the ground with an initial velocity $V_0$ can reach a point $(x, y)$ from two different angles, as long as the point $(x, y)$ is within the parabola of safety:$$y = \displaystyle{\frac{1}{2}\left(\frac{V_0^2}{g} - \frac{x^2}{\frac{V_0^2}{g}}\right)}$$

Homework Equations

Will edit if there is need of any.

The Attempt at a Solution

So far this was my progress:

First in regards to a), I went ahead to show that complementar angles resulted in the same range:$$\frac{V_0^2sin(2 \theta)}{g} = \frac{V_0^2sin(2 \times 90- \theta)}{g} = \frac{V_0^2sin(180 - \theta)}{g}\Longleftrightarrow sin \theta = sin(180 - \theta)$$From that, I continued:$$R = \frac{V_0^2sen(90-2 \delta)}{g} \Rightarrow Rg = V_0^2cos(2 \delta) \Rightarrow cos(2 \delta) = \frac{Rg}{V_0^2} \Rightarrow \delta = \frac{1}{2}cos^{-1}\left(\frac{Rg}{V_0^2}\right)$$My initial question is wether the first part of my demonstration was satisfactory, because for some reason I don't really think it was, considering I didn't include the $\theta = 45 \pm \delta$ on it...

And now for the b) part, I tried to consider the position equations:$$x(t)=(V_0cos \theta)t \\ y(t)=(V_0sin \theta)t - \frac{gt^2}{2}$$By eliminating t, we have:$$y = xtan \theta - \frac{g}{2} \frac{x^2}{V_0^2 cos^2 \theta}$$And now I don't know how can I continue the demonstration... x___x

I'd really appreciate some feedback on both of these problems!

 

[BvU]

I went ahead to show that complementar angles resulted in the same range

you may have lost something on the way. We can't tell if you don't post your steps so please guide us...

demonstration was satisfactory

It's not totally incorrect but somewhat trivial. It's not what the exercise composer meant .

 

[Bunny-chan]

you may have lost something on the way. We can't tell if you don't post your steps so please guide us...
It's not totally incorrect but somewhat trivial. It's not what the exercise composer meant .

Uhm. I'm confused now. What did I lose? D:
Doesn't that notation imply they are complementar?

And yes, I realized it's not really what was asked, but I don't know how should I proceed.

 

[Gear300]

The range R is given by V⁰₂sin(2θ)/g. The sin(2θ) solves part (a).

For part (b), the parabola of safety is the parabola centered at x = 0, whose height is the maximum height of ½V₀²/g and maximum range is ±V₀²/g. To show that any point under this envelope can occur twice, use

sec²θ = 1 + tan²θ

to reduce your bottom most equation to

αtan²θ + βtanθ + δ = 0,

where the coefficients are in terms of x and y. tanθ ranges as (-∞,∞), so all θ in the range (-π/2,π/2) are viable.

 

[BvU]

How do you solve $\frac{V_0^2\sin(2 \theta)}{g} = X$ ?
$$\frac{V_0^2\sin(2 \theta)}{g} \ne \frac{V_0^2\sin(\pi- \theta)}{g} $$Instead you want to solve
$$\frac{V_0^2\sin(2 \theta)}{g} = \frac{V_0^2\sin(\pi- 2\theta)}{g} $$
(note: use \sin instead of sin in $\LaTeX$)

 

[PeroK]

So far this was my progress:

First in regards to a), I went ahead to show that complementar angles resulted in the same range:

V20sin(2θ)g=V20sin(2×90−θ)g=V20sin(180−θ)g⟺sinθ=sin(180−θ)​

\frac{V_0^2sin(2 \theta)}{g} = \frac{V_0^2sin(2 \times 90- \theta)}{g} = \frac{V_0^2sin(180 - \theta)}{g}\Longleftrightarrow sin \theta = sin(180 - \theta)From that, I continued:

R=V20sen(90−2δ)g⇒Rg=V20cos(2δ)⇒cos(2δ)=RgV20⇒δ=12cos−1(RgV20)​

R = \frac{V_0^2sen(90-2 \delta)}{g} \Rightarrow Rg = V_0^2cos(2 \delta) \Rightarrow cos(2 \delta) = \frac{Rg}{V_0^2} \Rightarrow \delta = \frac{1}{2}cos^{-1}\left(\frac{Rg}{V_0^2}\right)My initial question is wether the first part of my demonstration was satisfactory, because for some reason I don't really think it was, considering I didn't include the θ=45±δ\theta = 45 \pm \delta on it...

Hmm! If you start from:

$R_{\pm} = \frac{V_0^2}{g} \sin(2\theta) = \frac{V_0^2}{g} \sin(90 \pm 2\delta)$

I suggest you can simply state that these are equal by symmetry of the $\sin$ function. And every value strictly between $0$ and $R_{max}$ is possible by continuity of the $\sin$ function.

PS Although, I see you have to calculate $\delta$ in any case.

 

[PeroK]

b) Generalize your previous result, showing that a projectile launched from the ground with an initial velocity V0V_0 can reach a point (x,y)(x, y) from two different angles, as long as the point (x,y)(x, y) is within the parabola of safety:

y=12⎛⎜⎝V20g−x2V20g⎞⎟⎠​

This is not strictly correct, as points vertically above the launch point can only be reached by one angle.

 

[Bunny-chan]

The range R is given by V⁰₂sin(2θ)/g. The sin(2θ) solves part (a).

For part (b), the parabola of safety is the parabola centered at x = 0, whose height is the maximum height of ½V₀²/g and maximum range is ±V₀²/g. To show that any point under this envelope can occur twice, use

sec²θ = 1 + tan²θ

to reduce your bottom most equation to

αtan²θ + βtanθ + δ = 0,

where the coefficients are in terms of x and y. tanθ ranges as (-∞,∞), so all θ in the range (-π/2,π/2) are viable.

I'm sorry, I didn't really follow what you mean x_x. But here's what I tried:

Noting $R_{max} = \frac{V_0^2}{g}$ and $r = \tan \theta$, we find a quadratic equation for $r$: $$y = rx- \left( \frac{1+r^2}{2R_{max}} \right) x^2$$Which we rewrite as$$x^2r^2-(2R_{max})r + (2R_{max}y + x^2) = 0$$The equation has two real solutions if$$(R_{max}x^2)-x^2(2R_{max}y + x^2) \geq 0$$or$$2R_{max}x^2 \left[ \frac{1}{2}\left(R_{max}-\frac{x^2}{R_{max}}\right)-y\right] \geq 0$$which means (x, y) is a point within the parabola. What do you guys think?

 

[Bunny-chan]

This is not strictly correct, as points vertically above the launch point can only be reached by one angle.

Yes, that's true, but still, I need to suggest a solution. D:

 

[PeroK]

Yes, that's true, but still, I need to suggest a solution. D:

You could set up a quadratic equation in $\tan \theta$ then use the discriminant to identify in terms of $x$ and $y$ when it has two solutions.

 

[Bunny-chan]

You could set up a quadratic equation in $\tan \theta$ then use the discriminant to identify in terms of $x$ and $y$ when it has two solutions.

Yes, I did that, check my above comments.
Do you think it's good?

 

[PeroK]

I'm sorry, I didn't really follow what you mean x_x. But here's what I tried:

Noting $R_{max} = \frac{V_0^2}{g}$ and $r = \tan \theta$, we find a quadratic equation for $r$: $$y = rx- \left( \frac{1+r^2}{2R_{max}} \right) x^2$$Which we rewrite as$$x^2r^2-(2R_{max})r + (2R_{max}y + x^2) = 0$$The equation has two real solutions if$$(R_{max}x^2)-x^2(2R_{max}y + x^2) \geq 0$$or$$2R_{max}x^2 \left[ \frac{1}{2}\left(R_{max}-\frac{x^2}{R_{max}}\right)-y\right] \geq 0$$which means (x, y) is a point within the parabola. What do you guys think?

I think it's right, but setting $R_{max}$ equal to half of what it should be was confusing.

 

[Bunny-chan]

I think it's right, but setting $R_{max}$ equal to half of what it should be was confusing.

Hmm... But why half? That's the value the exercise gives.

 

[PeroK]

Hmm... But why half? that's the value the exercise gives.

Yes, you're right.

 

[Bunny-chan]

Yes, you're right.

Thank you! By the way, what do you think in regards to a)? Is there a way I could make the answer more appropriate or is it fine?

 

[PeroK]

Re part a), I would prefer:

$R_{\pm} = \frac{V_0^2}{g} \sin(2\theta_{\pm}) = \frac{V_0^2}{g} \sin(90 \pm 2\delta) = R_{max} \cos(2\delta)$

 

[Bunny-chan]

Re part a), I would prefer:

$R_{\pm} = \frac{V_0^2}{g} \sin(2\theta_{\pm}) = \frac{V_0^2}{g} \sin(90 \pm 2\delta) = R_{max} \cos(2\delta)$

That seems good. Thanks a lot!