Projectile Motion Water hose Question

In summary: I got the answer.yep, I got the answer.In summary, the conversation discusses using a water hose to fill a large cylindrical storage tank. The hose shoots water at a 45° angle from the same level as the base of the tank, which is 6D away. The student attempts to use the equation Vx=Vt to solve the problem, but realizes it is not appropriate for this scenario. The correct equation, d=vtsin45+1/2at^2, is used to solve for the range of launch speeds that will allow the water to enter the tank. The student successfully solves the problem using this equation.
  • #1
jasonchiang97
72
2

Homework Statement


A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D . The hose shoots the water at 45 ∘ above the horizontal from the same level as the base of the tank and is a distance 6D away

Homework Equations


b1c5647476c66437d2183de53f091d77.png

8f817acb7ed26dd699a14912a0125f23.png

Vx=Vt
D=vit+1/2at^2

The Attempt at a Solution


I tried plugging in the values into the
8f817acb7ed26dd699a14912a0125f23.png
equation and I end up with√6dG<V<√√7dG
 
Physics news on Phys.org
  • #2
You should not apply equations without understanding the context in which they are valid. The equation you tried does not allow you to use the height of the tank, so it cannot be appropriate. When is it appropriate?
 
  • #3
haruspex said:
You should not apply equations without understanding the context in which they are valid. The equation you tried does not allow you to use the height of the tank, so it cannot be appropriate. When is it appropriate?

Right, okay I think I'm only allowed to use the equation when the object is landing at the same level ground as it started off on.
 
  • #4
jasonchiang97 said:
Right, okay I think I'm only allowed to use the equation when the object is landing at the same level ground as it started off on.
Right. Or, more precisely, when d is the distance to a point where it is at the same level it started at.
 
  • #5
jasonchiang97 said:

Homework Statement


A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D . The hose shoots the water at 45 ∘ above the horizontal from the same level as the base of the tank and is a distance 6D away
What is the question? What are you trying to find?
 
  • #6
jasonchiang97 said:

Homework Statement


A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D . The hose shoots the water at 45 ∘ above the horizontal from the same level as the base of the tank and is a distance 6D away.

For what range of launch speeds (v0) will the water enter the tank? Ignore air resistance, and express your answer in terms of D and g .

Homework Equations


b1c5647476c66437d2183de53f091d77.png

8f817acb7ed26dd699a14912a0125f23.png

Vx=Vt
D=vit+1/2at^2

The Attempt at a Solution


I tried plugging in the values into the
8f817acb7ed26dd699a14912a0125f23.png
equation and I end up with√6dG<V<√√7dG
 
  • #7
gneill said:
What is the question? What are you trying to find?

oops. Sorry I thought I posted that.
 
  • #8
Okay, so what are the scenarios for the water stream that you need to investigate? That is, what are the significant characteristics of the trajectory in each case?
 
  • #9
OK. From where is the hose 6D away? From the centre of the tank? Or from where the tank begins?
 
  • #10
[USER=569844]@navin[/USER] said:
OK. From where is the hose 6D away? From the centre of the tank? Or from where the tank begins?
Clever student's strategy: Unless the problem states a condition explicitly, choose the simplest situation to work with and declare it as an assumption. Just make sure that it's really undefined and not subtly implied somehow in the problem statement.
 
  • Like
Likes gracy
  • #11
You can't use dg = v^2(sin∆) here as this is equation gives us the maximum range of the projectile.
 
  • #12
[USER=569844]@navin[/USER] said:
You can't use dg = v^2(sin∆) here as this is equation gives us the maximum range of the projectile.
See post #3.
 
  • #13
jasonchiang97 said:
Right, okay I think I'm only allowed to use the equation when the object is landing at the same level ground as it started off on.
So what equation should you use here?
 
  • Like
Likes jasonchiang97
  • #14
haruspex said:
So what equation should you use here?

I used the 2D equations and plugged in t=Vx/Vcos45 into d=vtsin45+1/2at^2 and solved.

Thanks
 
  • #15
jasonchiang97 said:
I used the 2D equations and plugged in t=Vx/Vcos45 into d=vtsin45+1/2at^2 and solved.

Thanks
So you got the answer?
 
  • #16
haruspex said:
So you got the answer?

yep
 

1. How does the water flow in a projectile motion water hose?

The water flows in a parabolic path, similar to the path of a projectile. This is due to the force of gravity acting on the water, causing it to follow a curved trajectory.

2. What factors affect the range of a water hose in projectile motion?

The range of a water hose in projectile motion is affected by the initial velocity, angle of launch, and air resistance. These factors can increase or decrease the range of the water hose.

3. Can the range of a water hose be calculated using projectile motion equations?

Yes, the range of a water hose can be calculated using the projectile motion equations, as long as the initial velocity, angle of launch, and air resistance are known.

4. How does the range of a water hose change if the angle of launch is increased?

If the angle of launch is increased, the range of the water hose will also increase. This is because a higher angle will result in a higher initial velocity, which will cause the water to travel further before hitting the ground.

5. Is air resistance a significant factor in the trajectory of a water hose in projectile motion?

Air resistance does play a role in the trajectory of a water hose, but it is not a significant factor. This is because the mass and velocity of the water droplets are relatively low compared to other projectiles, and therefore air resistance has a minimal effect on their trajectory.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
19K
Replies
3
Views
3K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
Back
Top