# Homework Help: Projectiles launched at complementary angles

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1. Apr 19, 2017

### Bunny-chan

1. The problem statement, all variables and given/known data
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:$$y = \displaystyle{\frac{1}{2}\left(\frac{V_0^2}{g} - \frac{x^2}{\frac{V_0^2}{g}}\right)}$$
2. Relevant equations
Will edit if there is need of any.

3. 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!

2. Apr 19, 2017

### BvU

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 .

3. Apr 19, 2017

### Bunny-chan

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.

4. Apr 19, 2017

### 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.

5. Apr 19, 2017

### 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 $\LaTeX$)

6. Apr 19, 2017

### PeroK

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.

Last edited: Apr 19, 2017
7. Apr 19, 2017

### PeroK

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

8. Apr 20, 2017

### Bunny-chan

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?

9. Apr 20, 2017

### Bunny-chan

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

10. Apr 21, 2017

### PeroK

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.

11. Apr 21, 2017

### Bunny-chan

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

12. Apr 21, 2017

### PeroK

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

13. Apr 21, 2017

### Bunny-chan

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

Last edited: Apr 21, 2017
14. Apr 21, 2017

### PeroK

Yes, you're right.

15. Apr 21, 2017

### Bunny-chan

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?

16. Apr 21, 2017

### PeroK

Re part a), I would prefer:

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

17. Apr 21, 2017

### Bunny-chan

That seems good. Thanks a lot!