# Prove there exists no combination of speed and angles

1. Jan 30, 2014

### negation

1. The problem statement, all variables and given/known data

Two projectiles are launched simultaneously from the same point, with different launch speeds and angles. Show that no combination of speeds and angles will permit them to land simultaneously and at the same point.

3. The attempt at a solution

I have a rough idea. In short, it can be inferred that both projectiles will not land on the same point at the same time.
If my inference is flawed, how should I go about?

2. Jan 30, 2014

### Dick

What is your "rough idea" anyway? Just saying what you want to prove is true doesn't cut it.

3. Jan 30, 2014

### negation

In case of two projectile, I'll stipulate a constant range, x. Then determine the time taken for the projectile to land at that point. This is my inference.

4. Jan 30, 2014

### BvU

Ha, now you're in the sights of a heavyweight. I do agree with him, though. Also agree with your plan of approach. Wouldn't call it inference.

5. Jan 30, 2014

### Dick

You really have to say something more specific than that. How do you determine the time? Say something about how the horizontal and vertical distances vary with time.

6. Jan 30, 2014

### negation

Suppose projectile 1 has a launch angle of Θ + a with an initial velocity of vi + γ
and
projectile 2 has a launch angle of Θ and initial velocity of vi.

The stipulated range is a constant, x.

To determine the time, taken for projectile 1 to land at x:

t = x/ (vi + γ) cos (Θ+a)

To determine the time, taken for projectile 2 to land at x:

t = x/vi cos Θ

7. Jan 30, 2014

### Dick

Two projectiles launched with different speeds and different angles could have the same horizontal component of velocity. Then what?

8. Jan 30, 2014

### negation

I would suppose the vertical velocity is irrelevant in this question? Since the time taken for a projectile to achieve a displacement x can be determined by the horizontal velocity?

9. Jan 30, 2014

### Dick

It's very relevant. The projectiles could be at the same horizontal distance at the same time. Why don't you write out expressions for x(t) and y(t) in terms of the horizontal and vertical components of velocity?

10. Jan 30, 2014

### nil1996

I advice you to first try making equation of time and range of the projectiles.Then try to solve them and later compare them to given data.

11. Jan 30, 2014

### negation

But here:

t_full trajectory = 2vyi/g = 2vi sin Θ/g

x(t) = vi cos Θ . t

y(t) = y(0) + vi sin Θ. t - 0.5gt^2

12. Jan 30, 2014

### nil1996

Another way,
assume that they can reach the spot simultaneously.Then make equations of time and range.Solve them.You will get a contradictory result with the given data.

13. Jan 30, 2014

### negation

So the method is to use proof by reductio ad absurdum?

14. Jan 30, 2014

### nil1996

yes.We also call it Indirect Proof or Proof by contradiction

15. Jan 30, 2014

### negation

Well, I did something like a above.

Projectile 1 has an initial velocity of vi and launch angle of Θ
$Range, x = vi cos Θ . t$
tprojectile 1 = $x/vi cos Θ$

Projectile 2 has an initial velocity of $vi + \gamma$ and launch angle of
$Θ + \alpha$

tprojectile 2 = $x/ (vi +\gamma) cos (Θ+\alpha)$

If projectile 1 and projectile 2 takes the same time to cover range, x,
then tprojectile 1 = tprojectile 2

But tprojectile 1 =/ tprojectile 2

So, tprojectile 1 = tprojectile 2 is false.

Would this argument be a sufficient condition for the proof?

Last edited: Jan 30, 2014
16. Jan 30, 2014

### nil1996

:surprised We don't give Indirect proof like this!!!
you should make equations.Try mixing them and arrive at a conclusion which is contradictory with the given data.

17. Jan 30, 2014

### BvU

Agree with nil. From your t1 and t2 you haven't proven that they can't be equal. How can you be so sure there is no combination of alpha and gamma that makes t1=t2 ? Convince us!

18. Jan 30, 2014

### Dick

I think it's bit easier if you forget about angles for the moment. So for one projectile $$x_1(t)=v_{x1}t$$ $$y_1(t)=v_{y1}t-(1/2)gt^2$$ and for the other projectile $$x_2(t)=v_{x2}t$$ $$y_2(t)=v_{y2}t-(1/2)gt^2$$

19. Jan 31, 2014

### negation

I haven't forgotten about this problem. Just home from work and will be focusing on the other question first. I will give this problem further analysis tomorrow.

20. Jan 31, 2014

### Dick

No rush. It's actually pretty easy if you think of it that way.