# 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.

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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.

• topsquark
Homework Helper
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##.

[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 !

##\ ##

Last edited:
• DrClaude and topsquark
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##.

[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?

Homework Helper
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.

Homework Helper
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 • 