# Find angles such that the motion travels a specific distance

• BurpHa
In summary: I don't think it applies to everyone. You might want to add it in if you feel like it makes your summary more understandable.
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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BurpHa said:
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
Hello @BurpHa ,

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
BurpHa said:
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
BvU said:
Hello @BurpHa ,

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?

DrClaude said:
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 said:
Hi,

First of all, let me thank you for your effort for helping me.
BurpHa said:
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?

BurpHa said:
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 ?

##\ ##

BvU said:
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.

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

BvU said:
There is a ##\LaTeX## guide button to the lower left of the edit window...

View attachment 316195
Thank you!

BvU
Have fun !

BurpHa

## 1. How do you find the angles for a specific distance?

The angles can be found by using trigonometric functions such as sine, cosine, and tangent. These functions can be used to calculate the angles based on the distance and other known variables such as the velocity and time.

## 2. What is the importance of finding the angles for a specific distance?

Finding the angles is important because it helps determine the trajectory of the motion and ensures that the desired distance is achieved. It also allows for more accurate predictions and calculations in scientific experiments and real-world applications.

## 3. Can the angles be found using any other methods?

Yes, there are other methods such as using vector analysis or calculus to find the angles. However, trigonometric functions are often the most straightforward and efficient way to calculate the angles for a specific distance.

## 4. Are there any limitations to finding angles for a specific distance?

One limitation is that the calculations assume ideal conditions and do not account for external factors such as air resistance and friction. Additionally, the accuracy of the results may decrease with longer distances and higher velocities.

## 5. How can the angles be adjusted to achieve a different distance?

The angles can be adjusted by changing the initial velocity or time of the motion. By altering these variables, the angles can be recalculated to achieve a different distance. Additionally, changing the angle of launch can also affect the distance traveled.

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