Prove there exists no combination of speed and angles

[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.