Projectile Motion Water hose Question

[jasonchiang97]

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

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

The Attempt at a Solution

I tried plugging in the values into the

equation and I end up with√6dG<V<√√7dG

 

[haruspex]

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?

 

[jasonchiang97]

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.

 

[haruspex]

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.

 

[gneill]

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?

 

[jasonchiang97]

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

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

The Attempt at a Solution

I tried plugging in the values into the

equation and I end up with√6dG<V<√√7dG

 

[jasonchiang97]

What is the question? What are you trying to find?

oops. Sorry I thought I posted that.

 

[gneill]

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?

 

[@navin]

OK. From where is the hose 6D away? From the centre of the tank? Or from where the tank begins?

 

[gneill]

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.

 

[@navin]

You can't use dg = v^2(sin∆) here as this is equation gives us the maximum range of the projectile.

 

[haruspex]

You can't use dg = v^2(sin∆) here as this is equation gives us the maximum range of the projectile.

See post #3.

 

[haruspex]

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?

 

[jasonchiang97]

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

 

[haruspex]

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?

 

[jasonchiang97]

So you got the answer?

yep