Prove there exists no combination of speed and angles

[negation]

Homework Statement

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.

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?

 

[Dick]

Homework Statement

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.

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?

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

 

[negation]

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

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.

 

[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.

 

[Dick]

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.

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.

 

[negation]

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.

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

 

[Dick]

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

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

 

[negation]

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

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?

 

[Dick]

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?

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?

 

[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.

 

[negation]

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?

I could but it's in my head already.

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

 

[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.

 

[negation]

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.

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

 

[nil1996]

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

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

 

[negation]

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.

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
t_{projectile 1} = x/vi cos Θ

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

t_{projectile 2} = x/ (vi +\gamma) cos (Θ+\alpha)

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

But t_{projectile 1} =/ t_{projectile 2}

So, t_{projectile 1} = t_{projectile 2} is false.

Would this argument be a sufficient condition for the proof?

 

[nil1996]

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.

 

[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!

 

[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 $$x_2(t)=v_{x2}t$$ $$y_2(t)=v_{y2}t-(1/2)gt^2$$

 

[negation]

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 $$x_2(t)=v_{x2}t$$ $$y_2(t)=v_{y2}t-(1/2)gt^2$$

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.

 

[Dick]

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.

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

 

[negation]

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 $$x_2(t)=v_{x2}t$$ $$y_2(t)=v_{y2}t-(1/2)gt^2$$

Alright, so x(t) and y(t) are displacement as a function of time. Where do I take it from here? It's quite a stumbling block.

If the times are different for both particles, both particles would have different displacement given the same initial velocity.

 

[ehild]

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.

The times are the same (the projectiles launch and land simultaneously ), starting from the same point and launch at the same point, so both x and y are the same. What do these condition mean for the initial vertical and horizontal velocity components?


ehild

 

[D H]

Alright, so x(t) and y(t) are displacement as a function of time. Where do I take it from here? It's quite a stumbling block.

You don't care whether the trajectories the projectiles follow in the air differ. You only care whether these trajectories result in the projectiles hitting the ground at the same time and at the same point. This should tell you where to take those descriptions of the trajectories: Take them to the point in time and space where the projectiles hit the ground. You'll get a time of flight and displacement for one projectile, another time of flight and displacement for the other projectile. If the time of flight and displacement for the two projectiles are the same, what does this mean with regard to the initial velocity vector?

 

[negation]

The times are the same (the projectiles launch and land simultaneously ), starting from the same point and launch at the same point, so both x and y are the same. What do these condition mean for the initial vertical and horizontal velocity components?


ehild

If the time and position of launch are the same, and only the velocity varies, then, the only difference is the final displacement of the projectile from their origin.

 

[negation]

You don't care whether the trajectories the projectiles follow in the air differ. You only care whether these trajectories result in the projectiles hitting the ground at the same time and at the same point. This should tell you where to take those descriptions of the trajectories: Take them to the point in time and space where the projectiles hit the ground. You'll get a time of flight and displacement for one projectile, another time of flight and displacement for the other projectile. If the time of flight and displacement for the two projectiles are the same, what does this mean with regard to the initial velocity vector?

Then their initial velocity vector are equivalent.

 

[Dick]

Alright, so x(t) and y(t) are displacement as a function of time. Where do I take it from here? It's quite a stumbling block.

If the times are different for both particles, both particles would have different displacement given the same initial velocity.

Suppose T is the nonzero time when the two projectile arrive at the same point. That means $x_1(T)=x_2(T)$ and $y_1(T)=y_2(T)$. What does $x_1(T)=x_2(T)$ tell you about the relation between $v_{x1}$ and $v_{x2}$?

 

[D H]

Then their initial velocity vector are equivalent.

You can't just say this. You need to prove it. That is the point of this exercise.

 

[negation]

Suppose T is the nonzero time when the two projectile arrive at the same point. That means $x_1(T)=x_2(T)$ and $y_1(T)=y_2(T)$. What does $x_1(T)=x_2(T)$ tell you about the relation between $v_{x1}$ and $v_{x2}$?

Well, both displacement x and y are a function of time and their initial velocity.
If $x_1(T)=x_2(T)$ and $y_1(T)=y_2(T)$, then using backward deduction, it must be the case that either time, t, or initial velocity, vi, for both projectiles are equivalent. But if we assume time, t, for both projectile to be the same and achieving the same displacement, then their initial velocity must be the same.

But how do I demonstrate the proof?

 

[D H]

Well, both displacement x and y are a function of time and their initial velocity.
If $x_1(T)=x_2(T)$ and $y_1(T)=y_2(T)$, then using backward deduction, it must be the case that either time, t, or initial velocity, vi, for both projectiles are equivalent. But if we assume time, t, for both projectile to be the same and achieving the same displacement, then their initial velocity must be the same.

But how do I demonstrate the proof?

With algebra, not logic. This is not a logic problem.

Try answering Dick's question. He didn't ask about y (yet). He asked about x only: What does $x_1(T)=x_2(T)$ tell you about the relation between $v_{x1}$ and $v_{x2}$?

Dick practically gave you the answer to this question in post #18. You know $x_1(t)$ and $x_2(t)$ from that post. So equate them at time t=T. What does that tell you about $v_{x1}$ and $v_{x2}$?

 

[negation]

With algebra, not logic. This is not a logic problem.

Try answering Dick's question. He didn't ask about y (yet). He asked about x only: What does $x_1(T)=x_2(T)$ tell you about the relation between $v_{x1}$ and $v_{x2}$?

Dick practically gave you the answer to this question in post #18. You know $x_1(t)$ and $x_2(t)$ from that post. So equate them at time t=T. What does that tell you about $v_{x1}$ and $v_{x2}$?

x₁ = v_i1.t
x₁ = vi1 . t
t = x/vi1
x₁(t = x/vi1) = vi1(x/vi1)
x₁ = (vi1.x)/vi1

if x₁ = x₂
then
(vi1.x)/vi1 = x₂
(vi1.x)/vi1 = v_i2.t

 

[D H]

No! You are making this far too hard. This part is very simple!

You are given $x_1(t)=v_{x1}t$ and $x_2(t)=v_{x2}t$ and you are given that $x_1(T)=x_2(T)$ for some time $T$. Substitute $x_1(T)$ with $v_{x1}T$ and do the same with $x_2(T)$ (but now use the expression for $x_2$). The time $T$ presumably isn't zero (justify this!), so you can divide both sides of the equality by this quantity. What does that leave you with?

 

[negation]

No! You are making this far too hard. This part is very simple!

You are given $x_1(t)=v_{x1}t$ and $x_2(t)=v_{x2}t$ and you are given that $x_1(T)=x_2(T)$ for some time $T$. Substitute $x_1(T)$ with $v_{x1}T$ and do the same with $x_2(T)$ (but now use the expression for $x_2$). The time $T$ presumably isn't zero (justify this!), so you can divide both sides of the equality by this quantity. What does that leave you with?

x₁(t) = x₂

v_x1t₁ = v_x2t₂

t₂ = \frac{vx1.t1}{vx2}

v_x1t₁ = \frac{vx2.vx1.t1}{vx2}

 

[D H]

No! There is no $t_1$ or $t_2$. There is just a time $T$. You are making this way too hard!

 

[negation]

No! There is no $t_1$ or $t_2$. There is just a time $T$. You are making this way too hard!

What is this expression for x₂ you are referring to?
Isn't x₂ just vi2.t?

 

[negation]

No! There is no $t_1$ or $t_2$. There is just a time $T$. You are making this way too hard!

Let's try it again:

x₁(t) = x₂(t)

Given an arbitrary time, T, the displacement for x₁ can be determined.
And in equating V_x1.T = v_x2.t, the unknown v_x2 can be determined.

 

[D H]

Let's try it again:

x₁(t) = x₂(t)

Given an arbitrary time, T, the displacement for x₁ can be determined.
And in equating V_x1.T = v_x2.t, the unknown v_x2 can be determined.

No!

Why did you substitute t with T for x₁ but not for x₂? It's the same time for both.

 

[negation]

No!

Why did you substitute t with T for x₁ but not for x₂? It's the same time for both.

With all the smoke screen, it can be hard to cut through to the solution. But I think I'm seeing the general solution.


The objective is to prove v_x1 = v_x2

x₁(t) = x₂(t)

We ascribe an arbitrary time value projectile 1.
This can be expressed as: v_x1(T)

In this same arbitrary time, T, projectile 2 has a displacement of v_x2(T)

so, v_x1(T) = v_x2(T)

In order to determine v_x2,
[v_x1(T)]/ (T)

Therefore,

v_x1 = v_x2

 

[Dick]

With all the smoke screen, it can be hard to cut through to the solution. But I think I'm seeing the general solution.


The objective is to prove v_x1 = v_x2

x₁(t) = x₂(t)

We ascribe an arbitrary time value projectile 1.
This can be expressed as: v_x1(T)

In this same arbitrary time, T, projectile 2 has a displacement of v_x2(T)

so, v_x1(T) = v_x2(T)

In order to determine v_x2,
[v_x1(T)]/ (T)

Therefore,

v_x1 = v_x2

That's better. More algebra, less 'logic'. Just one detail, it's worth noting that argument only works if T is not zero. The projectiles actually are in the same place at the same time at t=0. Now what about y?

 

[negation]

That's better. More algebra, less 'logic'. Just one detail, it's worth noting that argument only works if T is not zero. The projectiles actually are in the same place at the same time at t=0. Now what about y?

My general solution is to find the time projectile 1 and 2 is in flight:

y₁(t) = y₂(t)
y₁(T) = y₂(T)
v_yi1 - 0.5gt^2 = v_yi2 - 0.5gt^2

T(v_yi1) - 0.5gt^2) = T(v_yi2)

If V_yi1 - 0.5gt = v_yi2 - 0.5gt

\ni

T₁ = 0 V 2v_yi1/g

T₂ = 0 V 2v_yi2/g

 

[Dick]

My general solution is to find the time projectile 1 and 2 is in flight:

y₁(t) = y₂(t)
y₁(T) = y₂(T)
v_yi1 - 0.5gt^2 = v_yi2 - 0.5gt^2

T(v_yi1) - 0.5gt^2) = T(v_yi2)

If V_yi1 - 0.5gt = v_yi2 - 0.5gt

\ni

T₁ = 0 V 2v_yi1/g

T₂ = 0 V 2v_yi2/g

That's pretty unclear and you've got some dubious algebra there, like T(v_yi1) - 0.5gt^2) = T(v_yi2). Can you try that again? There's only one T here. And the T=0 solution you don't really care much about. You already know the projectiles start in the same place at the same time.

 

[negation]

That's pretty unclear and you've got some dubious algebra there, like T(v_yi1) - 0.5gt^2) = T(v_yi2). Can you try that again? There's only one T here. And the T=0 solution you don't really care much about. You already know the projectiles start in the same place at the same time.

It was by carelessness that I had the 't' in place of 'T'.

y₁(t) = y₂(t)
y₁(T) = y₂(T)
v_yi1 - 0.5gT^2 = v_yi2 - 0.5gT^2

T(v_yi1 - 0.5gT) = T(v_yi2 - 0.5gT)

If V_yi1 - 0.5gT = v_yi2 - 0.5gT

\ni

T₁ = 0 V 2v_yi1/g

T₂ = 0 V 2v_yi2/g

In both projectile we shall ignore solution where t = 0.

 

[D H]

If V_yi1 - 0.5gT = v_yi2 - 0.5gT

Why are you missing the obvious so often? I'm going to break the rules of this site: Simply add 0.5gT to both sides of that expression and be done with it.

That you regularly miss the obvious, and that you call our supposed lack of help a "smoke screen" indicates to me that you are lacking some basic problem solving skills. You need some one-on-one, realtime help to help you learn those basic skills. If your school has a tutoring center you should go take advantage of it.

 

[Dick]

It was by carelessness that I had the 't' in place of 'T'.

y₁(t) = y₂(t)
y₁(T) = y₂(T)
v_yi1 - 0.5gT^2 = v_yi2 - 0.5gT^2

T(v_yi1 - 0.5gT) = T(v_yi2 - 0.5gT)

If V_yi1 - 0.5gT = v_yi2 - 0.5gT

\ni

T₁ = 0 V 2v_yi1/g

T₂ = 0 V 2v_yi2/g

In both projectile we shall ignore solution where t = 0.

You still have a $T_1$ and a $T_2$. Who cares about them, whatever they are. What you are trying to do is establish a relation between $v_{y1}$ and $v_{y2}$.

 

[negation]

Why are you missing the obvious so often? I'm going to break the rules of this site: Simply add 0.5gT to both sides of that expression and be done with it.

That you regularly miss the obvious, and that you call our supposed lack of help a "smoke screen" indicates to me that you are lacking some basic problem solving skills. You need some one-on-one, realtime help to help you learn those basic skills. If your school has a tutoring center you should go take advantage of it.

It's not the lack of problem solving skills.
Working on limited material and self- studying before the semester reopens implies a steep learning curve. And at times where my interpretation of the concept and mathematical reasoning is faulty, I do require time to deconstruct those faulty reasoning and replace then with the valid ones.
It's the holidays now by the way.

 

[Dick]

It's not the lack of problem solving skills.
Working on limited material and self- studying before the semester reopens implies a steep learning curve. And at times where my interpretation of the concept and mathematical reasoning is faulty, I do require time to deconstruct those faulty reasoning and replace then with the valid ones.
It's the holidays now by the way.

That's all fine, good luck on the self-study. You did really well in post 37 with the x-component. You set an "objective to prove vx1 = vx2" and did it. Your objective dealing with the y-component is to show vy1 = vy2. But you never quite got there. Once you have shown vx1 = vx2 and vy1 = vy2, you might want to step back use your mathematical reasoning to say what that says about the original question, which was actually asking about speeds and angles.

 

[negation]

That's all fine, good luck on the self-study. You did really well in post 37 with the x-component. You set an "objective to prove vx1 = vx2" and did it. Your objective dealing with the y-component is to show vy1 = vy2. But you never quite got there. Once you have shown vx1 = vx2 and vy1 = vy2, you might want to step back use your mathematical reasoning to say what that says about the original question, which was actually asking about speeds and angles.

x₁(t) = x₁(0) + v_i1cosΘ
x₁(T₁) = v_i1cosΘ.T₁
x₂(t) = x₂(0) + v_i2cosΘ
x₂(T₂) = v_i2cosΘ.T₂
x₁(T₁) = x₂(T₂)

∴ v_i1cosΘ.T₁ = v_i2cosΘ.T₂

p1: If T₁ = T₂ \ni v_i1cosΘ = v_i2cosΘ

y₁(t) = y₁(0) + v_i1sinΘ.t - 0.5gt²
y₁(T₁) = v_i1sinΘ.T - 0.5gT²
y₂(t) = y₂(0) + v_i2sinΘ.t - 0.5gt²
y₂(T₂) = v_i2sinΘ.T - 0.5gT²

∴ v_i1sinΘ - 0.5gT₁ = v_i2sinΘ - 0.5gT₂

p2: T₁ = T₂ \ni v_i1sinΘ = v_i2sinΘ

p3: \forall T₁ = T₂
\ni
[v_i1cosΘ = v_i2cosΘ] \wedge [v_i1sinΘ = v_i2sinΘ]

but if v_i1 = (v_i1+\alpha) and Θ = (Θ + \gamma)
\ni
x₁(T₁) = (v_i1+ \alpha) cos (Θ +\gamma).T₁

and if vi₂ = (v_i2 - \alpha) and Θ = (Θ - \gamma)
\ni
x₂(T₂) = (v_i2-\alpha) cos (Θ - \gamma)
while holding the horizontal displacement, v_icosΘ.T, true.
then
T₁ = [v_icosΘ.T]/ (v_i1+ \alpha) cos (Θ +\gamma)
\neq
T₂ = [v_icosΘ.T]/ (v_i2-\alpha) cos (Θ - \gamma)

∴ T₁ \neq T₂ if there exists a combination of angles and initial velocity.


Hence, no combination of angles and initial velocity exists to enable a projectile to simultaneously land at the same point and in time.