Find angles such that the motion travels a specific distance

[BurpHa]

Homework Statement

A fire hose held near the ground shoots water at a
speed of 6.5 m / s. At what angle(s) should the nozzle point
in order that the water land 2.5 m away?

Relevant Equations

X = V0 * t + 1/2at^2. t = 2V0y / 9.8

We know the time it takes the water complete the whole parabola is (sin(x) * 6.5 * 2) / 9.8.

So I come up with (sin(x) * 6.5 * 2) / 9.8 * cos(x) * 6.5 = 2.5, because the x component of the velocity is the same for the whole time.

But I get the results like these: x≈0.30929171+πn,1.26150461+πn.

I see the correct results are x=72.27889027+180n,17.72110972+180nx

When I use t = 2.5 / cos(x) * 6.5 and plug it in 0 = sin(x) * 6.5 * (2.5 / cos(x) * 6.5) - 4.9 * (2.5 / cos(x) * 6.5) ^ 2, I get the desired results.

I don't know what is not right in (sin(x) * 6.5 * 2) / 9.8 * cos(x) * 6.5 = 2.5 that gets me wrong.

 

[DrClaude]

But I get the results like these: x≈0.30929171+πn,1.26150461+πn.

I see the correct results are x=72.27889027+180n,17.72110972+180nx

The first above is in radians while the lower is in degrees.

 

[BvU]

Hello @BurpHa ,

$\qquad$ !​

For starters: don't use the symbol $x$ for angle AND for distance. It will cost you dearly sooner or later.
(especially if the exercise text already has a symbol $\theta_0$ for the angle !)

In your relevant equations you have a $V_0$. What is it ? Aha, after a long study, I deduce that the second one is Not $V_0\,y$ but $V_{0,y}$, a.k.a. $V_0\sin\theta_0$. Did I conclude correctly ?

And you know $t = 2\, V_0\sin\theta_0 /g$ is the time, which can be inserted in $x = V_{0,x} \, t$. Where $V_{0,x} = V_0 \cos\theta_0$.

One equation with one unknown.

Your results make no sense to me. Way too many digits and you want to restrict them to $0<\theta_0<{\pi\over 2}$

Re

I don't know what is not right in (sin(x) * 6.5 * 2) / 9.8 * cos(x) * 6.5 = 2.5 that gets me wrong.

Well, it can't be right, because the dimensions left and right are different !
[edit ] Oops, I think goofed here (imagined brackets )
[edit2]Actually, the expression is quite correct and indeed, $2\, V_0\sin\theta_0 /g \,V_0 \cos\theta_0$ is really 2.5 m.
So it must be going off the rails somewhere in your path to solve that for $\theta_0$.
What are your steps ?

[edit3]Ah, I see. What a mess. All because of chaotic symbol use!

You have $y = V_{0,y} t + {1\over 2} gt^2$ and solve for $y=0$, from which $t = 2V_{0,y}/g$.
For $x$ you have $x=V_{0,x}t = V_{0,x}\,2V_{0,y}/g = 2V_0^2/g \cos\theta_0\sin\theta_0$.

Solving for $\theta_0$: $\sin(2\theta_0) = xg/V_0^2 \Rightarrow 2\theta_0 = 0.618 \ \text {or} \ 2\theta_0 = \pi - 0.618$.

The former gives your $\theta_0 = 0.309$ which is correct (!) and the latter yields your $\theta_0 = 1.26$ which is also correct.

In short: well done and on to the next exercise !


$\$

 

[BurpHa]

Hello @BurpHa ,

$\qquad$ !​

For starters: don't use the symbol $x$ for angle AND for distance. It will cost you dearly sooner or later.
(especially if the exercise text already has a symbol $\theta_0$ for the angle !)

In your relevant equations you have a $V_0$. What is it ? Aha, after a long study, I deduce that the second one is Not $V_0\,y$ but $V_{0,y}$, a.k.a. $V_0\sin\theta_0$. Did I conclude correctly ?

And you know $t = 2\, V_0\sin\theta_0 /g$ is the time, which can be inserted in $x = V_{0,x} \, t$. Where $V_{0,x} = V_0 \cos\theta_0$.

One equation with one unknown.

Your results make no sense to me. Way too many digits and you want to restrict them to $0<\theta_0<{\pi\over 2}$

Re

Well, it can't be right, because the dimensions left and right are different !
[edit ] Oops, I think goofed here (imagined brackets )
[edit2]Actually, the expression is quite correct and indeed, $2\, V_0\sin\theta_0 /g \,V_0 \cos\theta_0$ is really 2.5 m.
So it must be going off the rails somewhere in your path to solve that for $\theta_0$.
What are your steps ?

[edit3]Ah, I see. What a mess. All because of chaotic symbol use!

You have $y = V_{0,y} t + {1\over 2} gt^2$ and solve for $y=0$, from which $t = 2V_{0,y}/g$.
For $x$ you have $x=V_{0,x}t = V_{0,x}\,2V_{0,y}/g = 2V_0^2/g \cos\theta_0\sin\theta_0$.

Solving for $\theta_0$: $\sin(2\theta_0) = xg/V_0^2 \Rightarrow 2\theta_0 = 0.618 \ \text {or} \ 2\theta_0 = \pi - 0.618$.

The former gives your $\theta_0 = 0.309$ which is correct (!) and the latter yields your $\theta_0 = 1.26$ which is also correct.

In short: well done and on to the next exercise !


$\$

Hi,

First of all, let me thank you for your effort.
However, why I only see a bunch of hashtags in your response? I suppose it creates some formatting, but I can't see it, and does it apply to other people too? Do I need to include hashtags in my equations later on?

 

[BurpHa]

The first above is in radians while the lower is in degrees.

Thank you for helping me out! I was telling my calculator to give results in degrees, but it keeps displaying those numbers ;)

 

[BurpHa]

Hi,

First of all, let me thank you for your effort for helping me.

However, why I only see a bunch of hashtags and weird use of slashing in your response? I suppose it creates some formatting, but I can't see it, and does it apply to other people too? Do I need to include hashtags and slashing in my equations later on?

 

[BvU]

why I only see a bunch of hashtags

That is strange. The MathJax gizmo is supposed to transmogrificate the hashtag-delimited $\LaTeX$ stuff to a presentation like

Do you get tohse hashtags in other threads too ?

$\$

 

[BurpHa]

That is strange. The MathJax gizmo is supposed to transmogrificate the hashtag-delimited $\LaTeX$ stuff to a presentation like

View attachment 316193

Do you get tohse hashtags in other threads too ?

$\$

Now I could see your beautifully formatted text after refreshing my browser several times. Could you tell me how could you do that? Actually, before posting this question, I was looking for the formatting options but could not find it.

 

[BvU]

There is a $\LaTeX$ guide button to the lower left of the edit window...

 

[BurpHa]

There is a $\LaTeX$ guide button to the lower left of the edit window...

View attachment 316195

Thank you!

 

[BvU]

Have fun !